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Vector Algebra — JEE Previous Year Questions

Every Vector Algebra question asked in JEE Main and JEE Advanced across the last 186 papers — 245 questions, each with its correct answer. Free to read, no account needed.

Questions

245

Papers it appeared in

173/186

Appearance rate

93%

All 245 Vector Algebra questions

Most recent papers first.

Q1·MathematicsSingle correctJEE Advanced 2026
For real numbers α,β,γ,δ\alpha, \beta, \gamma, \deltaα,β,γ,δ and μ\muμ, consider the matrix M=[α12−1213β13γδμ]M = \begin{bmatrix} \alpha & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{3}} & \beta & \frac{1}{\sqrt{3}} \\ \gamma & \delta & \mu \end{bmatrix}M=​α3​1​γ​2​1​βδ​−2​1​3​1​μ​​. Suppose that MMT=IMM^T = IMMT=I, where MTM^TMT is the transpose of the matrix MMM, and III is the 3×33 \times 33×3 identity matrix. Let u⃗=α i^+13 j^+γ k^,v⃗=12 i^+β j^+δ k^\vec{u} = \alpha\,\hat{i} + \frac{1}{\sqrt{3}}\,\hat{j} + \gamma\,\hat{k}, \vec{v} = \frac{1}{\sqrt{2}}\,\hat{i} + \beta\,\hat{j} + \delta\,\hat{k}u=αi^+3​1​j^​+γk^,v=2​1​i^+βj^​+δk^ and w⃗=−12 i^+13 j^+μ k^\vec{w} = -\frac{1}{\sqrt{2}}\,\hat{i} + \frac{1}{\sqrt{3}}\,\hat{j} + \mu\,\hat{k}w=−2​1​i^+3​1​j^​+μk^. Match each entry in List-I to the correct entry in List-II and choose the correct option.
List-IList-II
P.The value of γ2+δ2\gamma^2 + \delta^2γ2+δ2 is1.0
Q.If xu⃗+yv⃗+zw⃗=j^x\vec{u} + y\vec{v} + z\vec{w} = \hat{j}xu+yv+zw=j^​ for some real numbers x,yx, yx,y and zzz, then the value of xxx is2.1
R.The value of ∣u⃗⋅(v⃗×w⃗)∣|\vec{u} \cdot (\vec{v} \times \vec{w})|∣u⋅(v×w)∣ is3.12\frac{1}{\sqrt{2}}2​1​
S.The value of ∣u⃗×(v⃗×w⃗)∣|\vec{u} \times (\vec{v} \times \vec{w})|∣u×(v×w)∣ is4.13\frac{1}{\sqrt{3}}3​1​
5.56\frac{5}{6}65​
  1. (A)(P) → (5), (Q) → (4), (R) → (2), (S) → (1)
  2. (B)(P) → (4), (Q) → (5), (R) → (1), (S) → (2)
  3. (C)(P) → (5), (Q) → (3), (R) → (2), (S) → (1)
  4. (D)(P) → (5), (Q) → (4), (R) → (1), (S) → (2)

Correct answer: (A)

Step-by-step solution →
Q2·MathematicsSingle correctJEE Advanced 2026
Let a⃗,b⃗\vec{a}, \vec{b}a,b be two vectors, and let P,QP, QP,Q and RRR be the points with position vectors a⃗\vec{a}a, b⃗\vec{b}b and a⃗+b⃗\vec{a} + \vec{b}a+b, respectively, with respect to the origin OOO. If ∣a⃗+b⃗∣=21|\vec{a} + \vec{b}| = \sqrt{21}∣a+b∣=21​, ∣a⃗−b⃗∣=3|\vec{a} - \vec{b}| = 3∣a−b∣=3, and a⃗\vec{a}a and (a⃗−b⃗)(\vec{a} - \vec{b})(a−b) are perpendicular to each other, then the area of the triangle OPROPROPR is
  1. (A)3\sqrt{3}3​
  2. (B)32\dfrac{\sqrt{3}}{2}23​​
  3. (C)332\dfrac{3\sqrt{3}}{2}233​​
  4. (D)32\dfrac{3}{2}23​

Correct answer: (C)

Step-by-step solution →
Q3·MathematicsSingle correctJEE Main 2026
Let a⃗=4i^−j^+3k^\vec{a} = 4\hat{i} - \hat{j} + 3\hat{k}a=4i^−j^​+3k^, b⃗=10i^+2j^−k^\vec{b} = 10\hat{i} + 2\hat{j} - \hat{k}b=10i^+2j^​−k^ and a vector c⃗\vec{c}c be such that 2(a⃗×b⃗)+3(b⃗×c⃗)=0⃗2(\vec{a} \times \vec{b}) + 3(\vec{b} \times \vec{c}) = \vec{0}2(a×b)+3(b×c)=0. If a⃗⋅c⃗=15\vec{a} \cdot \vec{c} = 15a⋅c=15, then c⃗⋅(i^+j^−3k^)\vec{c} \cdot (\hat{i} + \hat{j} - 3\hat{k})c⋅(i^+j^​−3k^) is equal to:
  1. (A)−6-6−6
  2. (B)−5-5−5
  3. (C)−4-4−4
  4. (D)−3-3−3

Correct answer: (B)

Step-by-step solution →
Q4·MathematicsNumericalJEE Main 2026
If a⃗=i^+j^+k^\vec{a} = \hat{i} + \hat{j} + \hat{k}a=i^+j^​+k^, b⃗=j^−k^\vec{b} = \hat{j} - \hat{k}b=j^​−k^ and c⃗\vec{c}c be three vectors such that a⃗×c⃗=b⃗\vec{a} \times \vec{c} = \vec{b}a×c=b and a⃗⋅c⃗=3\vec{a} \cdot \vec{c} = 3a⋅c=3, then c⃗⋅(a⃗−2b⃗)\vec{c} \cdot (\vec{a} - 2\vec{b})c⋅(a−2b) is equal to ______.

Correct answer: 3

Step-by-step solution →
Q5·MathematicsSingle correctJEE Main 2026
Let a⃗=2i^+3j^+3k^\vec{a} = 2\hat{i} + 3\hat{j} + 3\hat{k}a=2i^+3j^​+3k^ and b⃗=6i^+3j^+3k^\vec{b} = 6\hat{i} + 3\hat{j} + 3\hat{k}b=6i^+3j^​+3k^. Then the square of the area of the triangle with adjacent sides determined by the vectors (2a⃗+3b⃗)(2\vec{a} + 3\vec{b})(2a+3b) and (a⃗−b⃗)(\vec{a} - \vec{b})(a−b) is :
  1. (A)450
  2. (B)900
  3. (C)1800
  4. (D)2400

Correct answer: (C)

Step-by-step solution →
Q6·MathematicsSingle correctJEE Main 2026
Let a⃗=7i^+j^−k^\vec{a} = \sqrt{7}\hat{i} + \hat{j} - \hat{k}a=7​i^+j^​−k^ and b⃗=j^+2k^\vec{b} = \hat{j} + 2\hat{k}b=j^​+2k^. If r⃗\vec{r}r is a vector such that r⃗×a⃗+a⃗×b⃗=0⃗\vec{r} \times \vec{a} + \vec{a} \times \vec{b} = \vec{0}r×a+a×b=0 and r⃗⋅a⃗=0\vec{r} \cdot \vec{a} = 0r⋅a=0, then ∣3r⃗∣2|3\vec{r}|^2∣3r∣2 is equal to:
  1. (A)44
  2. (B)54
  3. (C)86
  4. (D)132

Correct answer: (A)

Step-by-step solution →
Q7·MathematicsSingle correctJEE Main 2026
Let O be the origin, OP⃗=a⃗\vec{OP} = \vec{a}OP=a and OQ⃗=b⃗\vec{OQ} = \vec{b}OQ​=b. If R is the point on OP⃗\vec{OP}OP such that OP⃗=5OR⃗\vec{OP} = 5\vec{OR}OP=5OR, and M is the point such that OQ⃗=5RM⃗\vec{OQ} = 5\vec{RM}OQ​=5RM, then PM⃗\vec{PM}PM is equal to :
  1. (A)15(a⃗−4b⃗)\frac{1}{5}(\vec{a} - 4\vec{b})51​(a−4b)
  2. (B)15(b⃗−4a⃗)\frac{1}{5}(\vec{b} - 4\vec{a})51​(b−4a)
  3. (C)15(−a⃗+4b⃗)\frac{1}{5}(-\vec{a} + 4\vec{b})51​(−a+4b)
  4. (D)15(−b⃗+4a⃗)\frac{1}{5}(-\vec{b} + 4\vec{a})51​(−b+4a)

Correct answer: (B)

Step-by-step solution →
Q8·MathematicsSingle correctJEE Main 2026
Let u^\hat{u}u^ and v^\hat{v}v^ be unit vectors inclined at an acute angle such that ∣u^×v^∣=32|\hat{u} \times \hat{v}| = \frac{\sqrt{3}}{2}∣u^×v^∣=23​​. If A⃗=λu^+v^+(u^×v^)\vec{A} = \lambda\hat{u} + \hat{v} + (\hat{u} \times \hat{v})A=λu^+v^+(u^×v^), then λ is equal to:
  1. (A)43(A⃗⋅u^)−23(A⃗⋅v^)\frac{4}{3}(\vec{A} \cdot \hat{u}) - \frac{2}{3}(\vec{A} \cdot \hat{v})34​(A⋅u^)−32​(A⋅v^)
  2. (B)23(A⃗⋅u^)−13(A⃗⋅v^)\frac{2}{3}(\vec{A} \cdot \hat{u}) - \frac{1}{3}(\vec{A} \cdot \hat{v})32​(A⋅u^)−31​(A⋅v^)
  3. (C)43(A⃗⋅u^)+23(A⃗⋅v^)\frac{4}{3}(\vec{A} \cdot \hat{u}) + \frac{2}{3}(\vec{A} \cdot \hat{v})34​(A⋅u^)+32​(A⋅v^)
  4. (D)(A⃗⋅u^)−12(A⃗⋅v^)(\vec{A} \cdot \hat{u}) - \frac{1}{2}(\vec{A} \cdot \hat{v})(A⋅u^)−21​(A⋅v^)

Correct answer: (A)

Step-by-step solution →
Q9·MathematicsSingle correctJEE Main 2026
Two adjacent sides of a parallelogram PQRS are given by PQ⃗=j^+k^\vec{PQ} = \hat{j} + \hat{k}PQ​=j^​+k^ and PS⃗=i^−j^\vec{PS} = \hat{i} - \hat{j}PS=i^−j^​. If the side PS is rotated about the point P by an acute angle α\alphaα in the plane of the parallelogram so that it becomes perpendicular to the side PQ, then sin⁡2(5α2)−sin⁡2(α2)\sin^2\left(\frac{5\alpha}{2}\right) - \sin^2\left(\frac{\alpha}{2}\right)sin2(25α​)−sin2(2α​) is equal to:
  1. (A)12\frac{1}{2}21​
  2. (B)32\frac{\sqrt{3}}{2}23​​
  3. (C)34\frac{\sqrt{3}}{4}43​​
  4. (D)235\frac{2\sqrt{3}}{5}523​​

Correct answer: (B)

Step-by-step solution →
Q10·MathematicsSingle correctJEE Main 2026
Let the vectors a⃗=−i^+j^+3k^\vec{a} = -\hat{i} + \hat{j} + 3\hat{k}a=−i^+j^​+3k^ and b⃗=i^+3j^+k^\vec{b} = \hat{i} + 3\hat{j} + \hat{k}b=i^+3j^​+k^. For some λ,μ∈R\lambda, \mu \in \mathbb{R}λ,μ∈R, let c⃗=λa⃗+μb⃗\vec{c} = \lambda\vec{a} + \mu\vec{b}c=λa+μb. If c⃗⋅(3i^−6j^+2k^)=10\vec{c} \cdot (3\hat{i} - 6\hat{j} + 2\hat{k}) = 10c⋅(3i^−6j^​+2k^)=10 and c⃗⋅(i^+j^+k^)=−2\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = -2c⋅(i^+j^​+k^)=−2, then ∣c⃗∣2|\vec{c}|^2∣c∣2 is equal to:
  1. (A)888
  2. (B)121212
  3. (C)141414
  4. (D)151515

Correct answer: (B)

Step-by-step solution →
Q11·MathematicsSingle correctJEE Main 2026
If a⃗\vec{a}a and b⃗\vec{b}b are two vectors such that ∣a⃗∣=2|\vec{a}| = 2∣a∣=2 and ∣b⃗∣=3|\vec{b}| = 3∣b∣=3, then the maximum value of 3∣(3a⃗+2b⃗)∣+4∣(3a⃗−2b⃗)∣3\left|\left(3\vec{a} + 2\vec{b}\right)\right| + 4\left|\left(3\vec{a} - 2\vec{b}\right)\right|3​(3a+2b)​+4​(3a−2b)​ is :
  1. (A)30
  2. (B)36
  3. (C)60
  4. (D)72

Correct answer: (C)

Step-by-step solution →
Q12·MathematicsSingle correctJEE Main 2026
For three unit vectors a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c}a,b,c satisfying ∣a⃗−b⃗∣2+∣b⃗−c⃗∣2+∣c⃗−a⃗∣2=9\left|\vec{a}-\vec{b}\right|^{2} + \left|\vec{b}-\vec{c}\right|^{2} + \left|\vec{c}-\vec{a}\right|^{2} = 9​a−b​2+​b−c​2+∣c−a∣2=9 and ∣2a⃗+kb⃗+kc⃗∣=3\left|2\vec{a} + k\vec{b} + k\vec{c}\right| = 3​2a+kb+kc​=3, the positive value of k is :
  1. (A)3
  2. (B)6
  3. (C)4
  4. (D)5

Correct answer: (D)

Step-by-step solution →
Q13·MathematicsSingle correctJEE Main 2026
Let P be a point in the plane of the vector AB→=3i^+j^−k^\overrightarrow{AB} = 3\hat{i} + \hat{j} - \hat{k}AB=3i^+j^​−k^ and AC→=i^−j^+3k^\overrightarrow{AC} = \hat{i} - \hat{j} + 3\hat{k}AC=i^−j^​+3k^ such that P is equidistant from the lines AB and AC. If ∣AP→∣=52\left|\overrightarrow{AP}\right| = \frac{\sqrt{5}}{2}​AP​=25​​, then the area of the triangle ABP is :
  1. (A)2
  2. (B)32\frac{3}{2}23​
  3. (C)304\frac{\sqrt{30}}{4}430​​
  4. (D)264\frac{\sqrt{26}}{4}426​​

Correct answer: (C)

Step-by-step solution →
Q14·MathematicsSingle correctJEE Main 2026
Let a⃗=2i^+j^−2k^\vec{a}=2\hat{i}+\hat{j}-2\hat{k}a=2i^+j^​−2k^ , b⃗=i^+j^\vec{b}=\hat{i}+\hat{j}b=i^+j^​ and c⃗=a⃗×b⃗\vec{c}=\vec{a}\times\vec{b}c=a×b . Let d⃗\vec{d}d be a vector such that ∣d⃗−a⃗∣=11,∣c⃗×d⃗∣=3\left|\vec{d}-\vec{a}\right|=\sqrt{11},\left|\vec{c}\times\vec{d}\right|=3​d−a​=11​,​c×d​=3 and the angle between c⃗\vec{c}c and d⃗\vec{d}d is π4\frac{\pi}{4}4π​ . Then a⃗⋅d⃗\vec{a}\cdot\vec{d}a⋅d is equal to
  1. (A)111111
  2. (B)333
  3. (C)000
  4. (D)111

Correct answer: (C)

Step-by-step solution →
Q15·MathematicsSingle correctJEE Main 2026
Let a⃗=2i^−5j^+5k\vec{a} = 2\hat{i} - 5\hat{j} + 5ka=2i^−5j^​+5k and b⃗=i^−j^+3k\vec{b} = \hat{i} - \hat{j} + 3kb=i^−j^​+3k. If c⃗\vec{c}c is a vector such that 2(a⃗×c⃗)+3(b⃗×c⃗)=0⃗2(\vec{a} \times \vec{c}) + 3(\vec{b} \times \vec{c}) = \vec{0}2(a×c)+3(b×c)=0 and (a⃗−b⃗)⋅c⃗=−97(\vec{a} - \vec{b}) \cdot \vec{c} = -97(a−b)⋅c=−97, then ∣c⃗×k∣2\left| \vec{c} \times k \right|^{2}∣c×k∣2 is equal to
  1. (A)193
  2. (B)233
  3. (C)218
  4. (D)205

Correct answer: (C)

Step-by-step solution →
Q16·MathematicsSingle correctJEE Main 2026
Let a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c}a,b,c be three vectors such that a⃗×b⃗=2(a⃗×c⃗)\vec{a} \times \vec{b} = 2(\vec{a} \times \vec{c})a×b=2(a×c). If ∣a⃗∣=1,∣b⃗∣=4|\vec{a}| = 1, |\vec{b}| = 4∣a∣=1,∣b∣=4, ∣c⃗∣=2|\vec{c}| = 2∣c∣=2, and the angle between b⃗\vec{b}b and c⃗\vec{c}c is 60°, then ∣a⃗.c⃗∣|\vec{a}.\vec{c}|∣a.c∣ is :
  1. (A)2
  2. (B)4
  3. (C)0
  4. (D)1

Correct answer: (D)

Step-by-step solution →
Q17·MathematicsSingle correctJEE Main 2026
Let a⃗=−i^+j^+2k^\vec{a} = -\hat{i} + \hat{j} + 2\hat{k}a=−i^+j^​+2k^, b⃗=i^−j^−3k^\vec{b} = \hat{i} - \hat{j} - 3\hat{k}b=i^−j^​−3k^, c⃗=a⃗×b⃗\vec{c} = \vec{a} \times \vec{b}c=a×b and d⃗=c⃗×a⃗\vec{d} = \vec{c} \times \vec{a}d=c×a. Then (a⃗−b⃗)⋅d⃗\left(\vec{a} - \vec{b}\right) \cdot \vec{d}(a−b)⋅d is equal to :
  1. (A)4
  2. (B)-4
  3. (C)-2
  4. (D)2

Correct answer: (C)

Step-by-step solution →
Q18·MathematicsSingle correctJEE Main 2026
Let a⃗=i^−2j^+3k^\vec{a} = \hat{i} - 2\hat{j} + 3\hat{k}a=i^−2j^​+3k^, b⃗=2i^+j^−k^\vec{b} = 2\hat{i} + \hat{j} - \hat{k}b=2i^+j^​−k^, c⃗=λi^+j^+k^\vec{c} = \lambda\hat{i} + \hat{j} + \hat{k}c=λi^+j^​+k^ and v⃗=a⃗×b⃗\vec{v} = \vec{a} \times \vec{b}v=a×b. If v⃗.c⃗=11\vec{v}.\vec{c} = 11v.c=11 and the length of the projection of b⃗\vec{b}b on c⃗\vec{c}c is p, then 9p29p^29p2 is equal to :
  1. (A)9
  2. (B)6
  3. (C)4
  4. (D)12

Correct answer: (D)

Step-by-step solution →
Q19·MathematicsSingle correctJEE Main 2026
Let a⃗=2i^−j^+k^\vec{a} = 2\hat{i} - \hat{j} + \hat{k}a=2i^−j^​+k^ and b⃗=λj^+2k^\vec{b} = \lambda\hat{j} + 2\hat{k}b=λj^​+2k^, λ∈Z\lambda \in Zλ∈Z be two vectors, Let c⃗=a⃗×b⃗\vec{c} = \vec{a} \times \vec{b}c=a×b and d⃗\vec{d}d be a vector of magnitude 2 in yz-plane. If ∣c⃗∣=53|\vec{c}| = \sqrt{53}∣c∣=53​, then the maximum possible value of (c⃗⋅d⃗)2\left(\vec{c} \cdot \vec{d}\right)^{2}(c⋅d)2 is equal to :
  1. (A)26
  2. (B)104
  3. (C)208
  4. (D)52

Correct answer: (C)

Step-by-step solution →
Q20·MathematicsNumericalJEE Main 2026
Let a vector a⃗=2i^−j^+λk^\vec{a} = \sqrt{2}\hat{i} - \hat{j} + \lambda\hat{k}a=2​i^−j^​+λk^, λ>0\lambda > 0λ>0, make an obtuse angle with the vector b⃗=−λ2i^+42j^+42k^\vec{b} = -\lambda^2\hat{i} + 4\sqrt{2}\hat{j} + 4\sqrt{2}\hat{k}b=−λ2i^+42​j^​+42​k^ and an angle θ, π6<θ<π2\frac{\pi}{6} < \theta < \frac{\pi}{2}6π​<θ<2π​, with the positive z-axis. If the set of all possible values of λ is (α, β) – {γ}, then α + β + γ is equal to ______.

Correct answer: 5

Step-by-step solution →
Q21·MathematicsSingle correctJEE Main 2026
Let AB→=2i^+4j^−5k\overrightarrow{AB} = 2\hat{i} + 4\hat{j} - 5kAB=2i^+4j^​−5k and AD→=i^+2j^+λk\overrightarrow{AD} = \hat{i} + 2\hat{j} + \lambda kAD=i^+2j^​+λk, λ∈R\lambda \in \mathbb{R}λ∈R. Let the projection of the vector v⃗=i^+j^+k^\vec{v} = \hat{i} + \hat{j} + \hat{k}v=i^+j^​+k^ on the diagonal AC→\overrightarrow{AC}AC of the parallelogram ABCD be of length one unit. If α, β, where α > β, be the roots of the equation λ2x2−6λx+5=0\lambda^2 x^2 - 6\lambda x + 5 = 0λ2x2−6λx+5=0, then 2α − β is equal to
  1. (A)1
  2. (B)4
  3. (C)3
  4. (D)6

Correct answer: (C)

Step-by-step solution →
Q22·MathematicsSingle correctJEE Main 2026
Let a⃗=−i^+2j^+2k^\vec{a} = -\hat{i} + 2\hat{j} + 2\hat{k}a=−i^+2j^​+2k^, b⃗=8i^+7j^−3k^\vec{b} = 8\hat{i} + 7\hat{j} - 3\hat{k}b=8i^+7j^​−3k^ and c⃗\vec{c}c be a vector such that a⃗×c⃗=b⃗\vec{a} \times \vec{c} = \vec{b}a×c=b. If c⃗.(i^+j^+k^)=4\vec{c}.(\hat{i} + \hat{j} + \hat{k}) = 4c.(i^+j^​+k^)=4, then ∣a⃗+c⃗∣2|\vec{a} + \vec{c}|^{2}∣a+c∣2 is equal to :
  1. (A)33
  2. (B)30
  3. (C)35
  4. (D)27

Correct answer: (D)

Step-by-step solution →
Q23·MathematicsSingle correctJEE Main 2026
For a triangle ABC, let p⃗=BC→\vec{p} = \overrightarrow{BC}p​=BC, q⃗=CA→\vec{q} = \overrightarrow{CA}q​=CA and r⃗=BA→\vec{r} = \overrightarrow{BA}r=BA. If ∣p⃗∣=23|\vec{p}| = 2\sqrt{3}∣p​∣=23​, ∣q⃗∣=2|\vec{q}| = 2∣q​∣=2 and cos⁡θ=13\cos\theta = \frac{1}{\sqrt{3}}cosθ=3​1​, where θ\thetaθ is the angle between P⃗\vec{P}P and q⃗\vec{q}q​, then ∣p⃗×(q⃗−3r⃗)∣2+3∣r⃗∣2\left|\vec{p} \times \left(\vec{q} - 3\vec{r}\right)\right|^{2} + 3\left|\vec{r}\right|^{2}∣p​×(q​−3r)∣2+3∣r∣2 is equal to:
  1. (A)340
  2. (B)220
  3. (C)410
  4. (D)200

Correct answer: (D)

Step-by-step solution →
Q24·MathematicsNumericalJEE Advanced 2025
Consider the vectors x⃗=i^+2j^+3k^,y⃗=2i^+3j^+k^,\vec{x} = \hat{i} + 2\hat{j} + 3\hat{k}, \quad \vec{y} = 2\hat{i} + 3\hat{j} + \hat{k},x=i^+2j^​+3k^,y​=2i^+3j^​+k^, and z⃗=3i^+j^+2k^\vec{z} = 3\hat{i} + \hat{j} + 2\hat{k}z=3i^+j^​+2k^. For two distinct positive real numbers α and β, define X⃗=αx⃗+βy⃗−z⃗,Y⃗=αy⃗+βz⃗−x⃗,\vec{X} = \alpha\vec{x} + \beta\vec{y} - \vec{z}, \quad \vec{Y} = \alpha\vec{y} + \beta\vec{z} - \vec{x},X=αx+βy​−z,Y=αy​+βz−x, and Z⃗=αz⃗+βx⃗−y⃗\vec{Z} = \alpha\vec{z} + \beta\vec{x} - \vec{y}Z=αz+βx−y​. If the vectors X⃗,Y⃗\vec{X}, \vec{Y}X,Y, and Z⃗\vec{Z}Z lie in a plane, then the value of α + β − 3 is ______

Correct answer: -2

Step-by-step solution →
Q25·MathematicsSingle correctJEE Advanced 2025
Let w⃗=i^+j^−2k^\vec{w} = \hat{i} + \hat{j} - 2\hat{k}w=i^+j^​−2k^, and u⃗\vec{u}u and v⃗\vec{v}v be two vectors, such that u⃗×v⃗=w⃗\vec{u} \times \vec{v} = \vec{w}u×v=w and v⃗×w⃗=u⃗\vec{v} \times \vec{w} = \vec{u}v×w=u. Let α, β, γ, and t be real numbers such that u⃗=αi^+βj^+γk^\vec{u} = \alpha\hat{i} + \beta\hat{j} + \gamma\hat{k}u=αi^+βj^​+γk^, −tα+β+γ=0-t\alpha + \beta + \gamma = 0−tα+β+γ=0, α−tβ+γ=0\alpha - t\beta + \gamma = 0α−tβ+γ=0, and α+β−tγ=0\alpha + \beta - t\gamma = 0α+β−tγ=0. Match each entry in List-I to the correct entries in List-II and choose the correct option. The correct option is:
List-IList-II
P.∣v⃗∣2|\vec{v}|^2∣v∣2 is equal to1.0
Q.If α=3\alpha = \sqrt{3}α=3​, then γ2\gamma^2γ2 is equal to2.1
R.If α=3\alpha = \sqrt{3}α=3​, then (β+γ)2(\beta + \gamma)^2(β+γ)2 is equal to3.2
S.If α=2\alpha = \sqrt{2}α=2​, then t+3t + 3t+3 is equal to4.3
5.5
  1. (A)(P) → (2), (Q) → (1), (R) → (4), (S) → (5)
  2. (B)(P) → (2), (Q) → (4), (R) → (3), (S) → (5)
  3. (C)(P) → (2), (Q) → (1), (R) → (4), (S) → (3)
  4. (D)(P) → (5), (Q) → (4), (R) → (1), (S) → (3)

Correct answer: (A)

Step-by-step solution →
Q26·MathematicsNumericalJEE Advanced 2025
For any two points M and N in the XY –plane, let MN→\overrightarrow{MN}MN denote the vector from M to N, and 0⃗\vec{0}0 denote the zero vector. Let P, Q and R be three distinct points in the XY-plane. Let S be a point inside the triangle ΔPQR such that SP→+5SQ→+6SR→=0⃗\overrightarrow{SP} + 5\overrightarrow{SQ} + 6\overrightarrow{SR} = \vec{0}SP+5SQ​+6SR=0. Let E and F be the mid-points of the sides PR and QR, respectively. Then the value of length of the line segment EFlength of the line segment ES\frac{\text{length of the line segment EF}}{\text{length of the line segment ES}}length of the line segment ESlength of the line segment EF​ is __________.

Correct answer: 1.20

Step-by-step solution →
Q27·MathematicsSingle correctJEE Main 2025
Let a⃗=i^+2j^+k^\vec a=\hat i+2\hat j+\hat ka=i^+2j^​+k^ and b⃗=2i^+j^−k^\vec b=2\hat i+\hat j-\hat kb=2i^+j^​−k^. Let c^\hat cc^ be a unit vector in the plane of the vectors a⃗\vec aa and b⃗\vec bb and be perpendicular to a⃗\vec aa. Then such a vector c^\hat cc^ is:
  1. (A)12(j^−2k^)\dfrac{1}{\sqrt2}(\hat j-2\hat k)2​1​(j^​−2k^)
  2. (B)13(−i^−j^−k^)\dfrac{1}{\sqrt3}(-\hat i-\hat j-\hat k)3​1​(−i^−j^​−k^)
  3. (C)13(i^−j^+k^)\dfrac{1}{\sqrt3}(\hat i-\hat j+\hat k)3​1​(i^−j^​+k^)
  4. (D)12(−i^+k^)\dfrac{1}{\sqrt2}(-\hat i+\hat k)2​1​(−i^+k^)

Correct answer: (D)

Step-by-step solution →
Q28·MathematicsSingle correctJEE Main 2025
Let a⃗\vec{a}a and b⃗\vec{b}b be the vectors of the same magnitude such that ∣a⃗+b⃗∣+∣a⃗−b⃗∣∣a⃗+b⃗∣−∣a⃗−b⃗∣=2+1\dfrac{|\vec{a}+\vec{b}|+|\vec{a}-\vec{b}|}{|\vec{a}+\vec{b}|-|\vec{a}-\vec{b}|}=\sqrt{2}+1∣a+b∣−∣a−b∣∣a+b∣+∣a−b∣​=2​+1. Then ∣a⃗+b⃗∣2∣a⃗∣2\dfrac{|\vec{a}+\vec{b}|^2}{|\vec{a}|^2}∣a∣2∣a+b∣2​ is:
  1. (A)2+422+4\sqrt{2}2+42​
  2. (B)1+21+\sqrt{2}1+2​
  3. (C)2+22+\sqrt{2}2+2​
  4. (D)4+224+2\sqrt{2}4+22​

Correct answer: (C)

Step-by-step solution →
Q29·MathematicsSingle correctJEE Main 2025
Let the angle θ\thetaθ, 0<θ<π20<\theta<\dfrac{\pi}{2}0<θ<2π​ between two unit vectors a^\hat{a}a^ and b^\hat{b}b^ be sin⁡−1(659)\sin^{-1}\left(\dfrac{\sqrt{65}}{9}\right)sin−1(965​​). If the vector c⃗=3a^+6b^+9(a^×b^)\vec{c}=3\hat{a}+6\hat{b}+9(\hat{a}\times\hat{b})c=3a^+6b^+9(a^×b^), then the value of 9(c⃗⋅a^)−3(c⃗⋅b^)9(\vec{c}\cdot\hat{a})-3(\vec{c}\cdot\hat{b})9(c⋅a^)−3(c⋅b^) is
  1. (A)31
  2. (B)27
  3. (C)29
  4. (D)24

Correct answer: (C)

Step-by-step solution →
Q30·MathematicsIntegerJEE Main 2025
Let the three sides of a triangle ABC be given by the vectors 2i^−j^+k^2\hat{i}-\hat{j}+\hat{k}2i^−j^​+k^, i^−3j^−5k^\hat{i}-3\hat{j}-5\hat{k}i^−3j^​−5k^ and 3i^−4j^−4k^3\hat{i}-4\hat{j}-4\hat{k}3i^−4j^​−4k^. Let G be the centroid of the triangle ABC. Then 6(∣AG∣2+∣BG∣2+∣CG∣2)6\left(|AG|^2+|BG|^2+|CG|^2\right)6(∣AG∣2+∣BG∣2+∣CG∣2) is equal to ______.

Correct answer: 164

Step-by-step solution →
Q31·MathematicsSingle correctJEE Main 2025
Consider two vectors u⃗=3i^−j^\vec{u}=3\hat{i}-\hat{j}u=3i^−j^​ and v⃗=2i^+j^−λk^\vec{v}=2\hat{i}+\hat{j}-\lambda\hat{k}v=2i^+j^​−λk^, λ>0\lambda>0λ>0. The angle between them is given by cos⁡−1(527)\cos^{-1}\left(\dfrac{\sqrt{5}}{2\sqrt{7}}\right)cos−1(27​5​​). Let v⃗=v1⃗+v2⃗\vec{v}=\vec{v_1}+\vec{v_2}v=v1​​+v2​​, where v1⃗\vec{v_1}v1​​ is parallel to u⃗\vec{u}u and v2⃗\vec{v_2}v2​​ is perpendicular to u⃗\vec{u}u. Then the value of ∣v1⃗∣2+∣v2⃗∣2|\vec{v_1}|^2+|\vec{v_2}|^2∣v1​​∣2+∣v2​​∣2 is equal to:
  1. (A)232\dfrac{23}{2}223​
  2. (B)14
  3. (C)252\dfrac{25}{2}225​
  4. (D)10

Correct answer: (B)

Step-by-step solution →
Q32·MathematicsIntegerJEE Main 2025
Let a⃗=i^+j^+k^\vec a=\hat i+\hat j+\hat ka=i^+j^​+k^, b⃗=3i^+2j^−k^\vec b=3\hat i+2\hat j-\hat kb=3i^+2j^​−k^, c⃗=λj^+μk^\vec c=\lambda\hat j+\mu\hat kc=λj^​+μk^ and d^\hat dd^ be a unit vector such that a⃗×d^=b⃗×d^\vec a\times\hat d=\vec b\times\hat da×d^=b×d^ and c⃗⋅d^=1\vec c\cdot\hat d=1c⋅d^=1. If c⃗\vec cc is perpendicular to a⃗\vec aa, then ∣3λd^+μc⃗∣2\left|3\lambda\hat d+\mu\vec c\right|^2​3λd^+μc​2 is equal to __________.

Correct answer: 5

Step-by-step solution →
Q33·MathematicsIntegerJEE Main 2025
Let a⃗=i^+2j^+k^\vec a=\hat i+2\hat j+\hat ka=i^+2j^​+k^, b⃗=3i^−3j^+3k^\vec b=3\hat i-3\hat j+3\hat kb=3i^−3j^​+3k^, c⃗=2i^−j^+2k^\vec c=2\hat i-\hat j+2\hat kc=2i^−j^​+2k^ and d⃗\vec dd be a vector such that b⃗×d⃗=c⃗×d⃗\vec b\times\vec d=\vec c\times\vec db×d=c×d and a⃗⋅d⃗=4\vec a\cdot\vec d=4a⋅d=4. Then ∣a⃗×d⃗∣2\left|\vec a\times\vec d\right|^2​a×d​2 is equal to __________.

Correct answer: 128

Step-by-step solution →
Q34·MathematicsSingle correctJEE Main 2025
If a⃗\vec{a}a is nonzero vector such that its projections on the vectors 2i^−j^+2k^2\hat{i}-\hat{j}+2\hat{k}2i^−j^​+2k^, i^+2j^−2k^\hat{i}+2\hat{j}-2\hat{k}i^+2j^​−2k^ and k^\hat{k}k^ are equal, then a unit vector along a⃗\vec{a}a is:
  1. (A)1155(−7i^+9j^+5k^)\dfrac{1}{\sqrt{155}}(-7\hat{i}+9\hat{j}+5\hat{k})155​1​(−7i^+9j^​+5k^)
  2. (B)1155(−7i^+9j^−5k^)\dfrac{1}{\sqrt{155}}(-7\hat{i}+9\hat{j}-5\hat{k})155​1​(−7i^+9j^​−5k^)
  3. (C)1155(7i^+9j^+5k^)\dfrac{1}{\sqrt{155}}(7\hat{i}+9\hat{j}+5\hat{k})155​1​(7i^+9j^​+5k^)
  4. (D)1155(7i^+9j^−5k^)\dfrac{1}{\sqrt{155}}(7\hat{i}+9\hat{j}-5\hat{k})155​1​(7i^+9j^​−5k^)

Correct answer: (C)

Step-by-step solution →
Q35·MathematicsSingle correctJEE Main 2025
Let a⃗=2i^−3j^+k^\vec{a}=2\hat{i}-3\hat{j}+\hat{k}a=2i^−3j^​+k^, b⃗=3i^+2j^+5k^\vec{b}=3\hat{i}+2\hat{j}+5\hat{k}b=3i^+2j^​+5k^ and a vector c⃗\vec{c}c be such that (a⃗−c⃗)×b⃗=−18i^−3j^+12k^(\vec{a}-\vec{c})\times\vec{b}=-18\hat{i}-3\hat{j}+12\hat{k}(a−c)×b=−18i^−3j^​+12k^ and a⃗⋅c⃗=3\vec{a}\cdot\vec{c}=3a⋅c=3. If b⃗×c⃗=d⃗\vec{b}\times\vec{c}=\vec{d}b×c=d, then ∣a⃗⋅d⃗∣|\vec{a}\cdot\vec{d}|∣a⋅d∣ is equal to:
  1. (A)18
  2. (B)12
  3. (C)9
  4. (D)15

Correct answer: (D)

Step-by-step solution →
Q36·MathematicsSingle correctJEE Main 2025
Let ABCDABCDABCD be a tetrahedron such that the edges ABABAB, ACACAC and ADADAD are mutually perpendicular. Let the areas of the triangles ABCABCABC, ACDACDACD and ADBADBADB be 5, 6 and 7 square units respectively. Then the area (in square units) of the △BCD\triangle BCD△BCD is equal to:
  1. (A)340\sqrt{340}340​
  2. (B)12
  3. (C)110\sqrt{110}110​
  4. (D)737\sqrt{3}73​

Correct answer: (C)

Step-by-step solution →
Q37·MathematicsSingle correctJEE Main 2025
Let a⃗=2i^−j^+3k^\vec{a}=2\hat{i}-\hat{j}+3\hat{k}a=2i^−j^​+3k^, b⃗=3i^−5j^+k^\vec{b}=3\hat{i}-5\hat{j}+\hat{k}b=3i^−5j^​+k^ and c⃗\vec{c}c be a vector such that a⃗×c⃗=c⃗×b⃗\vec{a}\times\vec{c}=\vec{c}\times\vec{b}a×c=c×b and (a⃗+c⃗)⋅(b⃗+c⃗)=168(\vec{a}+\vec{c})\cdot(\vec{b}+\vec{c})=168(a+c)⋅(b+c)=168. Then the maximum value of ∣c⃗∣2|\vec{c}|^2∣c∣2 is:
  1. (A)77
  2. (B)462
  3. (C)308
  4. (D)154

Correct answer: (C)

Step-by-step solution →
Q38·MathematicsSingle correctJEE Main 2025
Let a^\hat{a}a^ be a unit vector perpendicular to the vectors b⃗=i^−2j^+3k^\vec{b}=\hat{i}-2\hat{j}+3\hat{k}b=i^−2j^​+3k^ and c⃗=2i^+3j^−k^\vec{c}=2\hat{i}+3\hat{j}-\hat{k}c=2i^+3j^​−k^, and makes an angle of cos⁡−1(−13)\cos^{-1}\left(-\dfrac{1}{3}\right)cos−1(−31​) with the vector i^+j^+k^\hat{i}+\hat{j}+\hat{k}i^+j^​+k^. If a^\hat{a}a^ makes an angle of π3\dfrac{\pi}{3}3π​ with the vector i^+αj^+k^\hat{i}+\alpha\hat{j}+\hat{k}i^+αj^​+k^, then the value of α\alphaα is:
  1. (A)−3-\sqrt{3}−3​
  2. (B)6\sqrt{6}6​
  3. (C)−6-\sqrt{6}−6​
  4. (D)3\sqrt{3}3​

Correct answer: (C)

Step-by-step solution →
Q39·MathematicsSingle correctJEE Main 2025
Let a⃗=i^+2j^+k^\vec{a}=\hat{i}+2\hat{j}+\hat{k}a=i^+2j^​+k^ and b⃗=2i^+7j^+3k^\vec{b}=2\hat{i}+7\hat{j}+3\hat{k}b=2i^+7j^​+3k^. Let L1:r⃗=(−i^+2j^+k^)+λa⃗L_1:\vec{r}=(-\hat{i}+2\hat{j}+\hat{k})+\lambda\vec{a}L1​:r=(−i^+2j^​+k^)+λa, λ∈R\lambda\in Rλ∈R and L2:r⃗=(j^+k^)+μb⃗L_2:\vec{r}=(\hat{j}+\hat{k})+\mu\vec{b}L2​:r=(j^​+k^)+μb, μ∈R\mu\in Rμ∈R be two lines. If the line L3L_3L3​ passes through the point of intersection of L1L_1L1​ and L2L_2L2​, and is parallel to a⃗+b⃗\vec{a}+\vec{b}a+b, then L3L_3L3​ passes through the point:
  1. (A)(8,26,12)(8, 26, 12)(8,26,12)
  2. (B)(2,8,5)(2, 8, 5)(2,8,5)
  3. (C)(−1,−1,1)(-1, -1, 1)(−1,−1,1)
  4. (D)(5,17,4)(5, 17, 4)(5,17,4)

Correct answer: (A)

Step-by-step solution →
Q40·MathematicsSingle correctJEE Main 2025
If the components of a⃗=αi^+βj^+γk^\vec{a}=\alpha\hat{i}+\beta\hat{j}+\gamma\hat{k}a=αi^+βj^​+γk^ along and perpendicular to b⃗=3i^+j^−k^\vec{b}=3\hat{i}+\hat{j}-\hat{k}b=3i^+j^​−k^ respectively, are 1611(3i^+j^−k^)\frac{16}{11}(3\hat{i}+\hat{j}-\hat{k})1116​(3i^+j^​−k^) and 111(−4i^−5j^−17k^)\frac{1}{11}(-4\hat{i}-5\hat{j}-17\hat{k})111​(−4i^−5j^​−17k^), then α2+β2+γ2\alpha^2+\beta^2+\gamma^2α2+β2+γ2 is equal to:
  1. (A)23
  2. (B)18
  3. (C)16
  4. (D)26

Correct answer: (D)

Step-by-step solution →
Q41·MathematicsIntegerJEE Main 2025
Let a⃗=i^+j^+k^\vec{a}=\hat{i}+\hat{j}+\hat{k}a=i^+j^​+k^, b⃗=2i^+2j^+k^\vec{b}=2\hat{i}+2\hat{j}+\hat{k}b=2i^+2j^​+k^ and d⃗=a⃗×b⃗\vec{d}=\vec{a}\times\vec{b}d=a×b. If c⃗\vec{c}c is a vector such that a⃗⋅c⃗=∣c⃗∣\vec{a}\cdot\vec{c}=|\vec{c}|a⋅c=∣c∣, ∣c⃗−2d⃗∣=8|\vec{c}-2\vec{d}|=8∣c−2d∣=8 and the angle between d⃗\vec{d}d and c⃗\vec{c}c is π4\frac{\pi}{4}4π​, then ∣10−3b⃗⋅c⃗∣+∣d⃗×c⃗∣2|10-3\vec{b}\cdot\vec{c}|+|\vec{d}\times\vec{c}|^2∣10−3b⋅c∣+∣d×c∣2 is equal to ____.

Correct answer: 6

Step-by-step solution →
Q42·MathematicsSingle correctJEE Main 2025
Let A, B, C be three points in xy-plane, whose position vector are given by 3i^+j^\sqrt{3}\hat{i}+\hat{j}3​i^+j^​, i^+3j^\hat{i}+\sqrt{3}\hat{j}i^+3​j^​ and ai^+(1−a)j^a\hat{i}+(1-a)\hat{j}ai^+(1−a)j^​ respectively with respect to the origin O. If the distance of the point C from the point bisecting the angle between the vectors OA⃗\vec{OA}OA and OB⃗\vec{OB}OB is 92\frac{9}{\sqrt{2}}2​9​, then the sum of all the possible values of a is:
  1. (A)1
  2. (B)92\frac{9}{2}29​
  3. (C)0
  4. (D)2

Correct answer: (A)

Step-by-step solution →
Q43·MathematicsSingle correctJEE Main 2025
Let a⃗=i^+2j^+3k^\vec a=\hat i+2\hat j+3\hat ka=i^+2j^​+3k^, b⃗=3i^+j^−k^\vec b=3\hat i+\hat j-\hat kb=3i^+j^​−k^ and c⃗\vec cc be three vectors such that c⃗\vec cc is coplanar with a⃗\vec aa and b⃗\vec bb. If the vector c⃗\vec cc is perpendicular to b⃗\vec bb and a⃗⋅c⃗=5\vec a\cdot\vec c=5a⋅c=5, then ∣c⃗∣|\vec c|∣c∣ is equal to
  1. (A)132\dfrac{1}{3\sqrt2}32​1​
  2. (B)181818
  3. (C)161616
  4. (D)116\sqrt{\dfrac{11}{6}}611​​

Correct answer: (D)

Step-by-step solution →
Q44·MathematicsSingle correctJEE Main 2025
Let the position vectors of three vertices of a triangle be 4p⃗+q⃗−3r⃗4\vec p+\vec q-3\vec r4p​+q​−3r, −5p⃗+q⃗+2r⃗-5\vec p+\vec q+2\vec r−5p​+q​+2r and 2p⃗−q⃗+2r⃗2\vec p-\vec q+2\vec r2p​−q​+2r. If the position vectors of the orthocenter and the circumcenter of the triangle are p⃗+q⃗+r⃗4\dfrac{\vec p+\vec q+\vec r}{4}4p​+q​+r​ and αp⃗+βq⃗+γr⃗\alpha\vec p+\beta\vec q+\gamma\vec rαp​+βq​+γr respectively, then α+2β+5γ\alpha+2\beta+5\gammaα+2β+5γ is equal to:
  1. (A)3
  2. (B)1
  3. (C)6
  4. (D)4

Correct answer: (A)

Step-by-step solution →
Q45·MathematicsSingle correctJEE Main 2025
Let a⃗=3i^−j^+2k^\vec a=3\hat i-\hat j+2\hat ka=3i^−j^​+2k^, b⃗=a⃗×(i^−3k^)\vec b=\vec a\times(\hat i-3\hat k)b=a×(i^−3k^) and c⃗=b⃗×k^\vec c=\vec b\times\hat kc=b×k^. Then the projection of c⃗−2j^\vec c-2\hat jc−2j^​ on a⃗\vec aa is:
  1. (A)373\sqrt737​
  2. (B)14\sqrt{14}14​
  3. (C)2142\sqrt{14}214​
  4. (D)272\sqrt727​

Correct answer: (C)

Step-by-step solution →
Q46·MathematicsSingle correctJEE Main 2025
Let the point AAA divide the line segment joining the points P(−1,−1,2)P(-1,-1,2)P(−1,−1,2) and Q(5,5,10)Q(5,5,10)Q(5,5,10) internally in the ratio r:1r:1r:1 (r>0)(r>0)(r>0). If OOO is the origin and (OQ→⋅OA→−15 ∣OP→×OA→∣2)=10\left(\overrightarrow{OQ}\cdot\overrightarrow{OA}-\tfrac{1}{5}\,|\overrightarrow{OP}\times\overrightarrow{OA}|^2\right)=10(OQ​⋅OA−51​∣OP×OA∣2)=10, then the value of rrr is :
  1. (A)14
  2. (B)3
  3. (C)7\sqrt{7}7​
  4. (D)7

Correct answer: (D)

Step-by-step solution →
Q47·MathematicsSingle correctJEE Main 2025
Let the position vectors of the vertices AAA, BBB and CCC of a tetrahedron ABCDABCDABCD be i^+2j^+k^\hat i+2\hat j+\hat ki^+2j^​+k^, i^+3j^−2k^\hat i+3\hat j-2\hat ki^+3j^​−2k^ and 2i^+j^−k^2\hat i+\hat j-\hat k2i^+j^​−k^ respectively. The altitude from the vertex DDD to the opposite face ABCABCABC meets the median line segment through AAA of the triangle ABCABCABC at the point EEE. If the length of ADADAD is 1103\dfrac{\sqrt{110}}{3}3110​​ and the volume of the tetrahedron is 80562\dfrac{\sqrt{805}}{6\sqrt2}62​805​​, then the position vector of EEE is
  1. (A)12(i^+4j^+7k^)\dfrac{1}{2}(\hat i+4\hat j+7\hat k)21​(i^+4j^​+7k^)
  2. (B)112(7i^+4j^+3k^)\dfrac{1}{12}(7\hat i+4\hat j+3\hat k)121​(7i^+4j^​+3k^)
  3. (C)16(12i^+12j^+k^)\dfrac{1}{6}(12\hat i+12\hat j+\hat k)61​(12i^+12j^​+k^)
  4. (D)16(7i^+12j^+k^)\dfrac{1}{6}(7\hat i+12\hat j+\hat k)61​(7i^+12j^​+k^)

Correct answer: (D)

Step-by-step solution →
Q48·MathematicsSingle correctJEE Main 2025
Let the arc ACACAC of a circle subtend a right angle at the centre OOO. If the point BBB on the arc ACACAC divides the arc ACACAC such that length of arc ABlength of arc BC=15\dfrac{\text{length of arc }AB}{\text{length of arc }BC}=\dfrac{1}{5}length of arc BClength of arc AB​=51​, and OC→=α OA→+β OB→\overrightarrow{OC}=\alpha\,\overrightarrow{OA}+\beta\,\overrightarrow{OB}OC=αOA+βOB, then α+2(3−1)β\alpha+\sqrt2(\sqrt3-1)\betaα+2​(3​−1)β is equal to
  1. (A)2−32-\sqrt32−3​
  2. (B)232\sqrt323​
  3. (C)535\sqrt353​
  4. (D)2+32+\sqrt32+3​

Correct answer: (A)

Step-by-step solution →
Q49·MathematicsSingle correctJEE Main 2025
Let a⃗\vec{a}a and b⃗\vec{b}b be two unit vectors such that the angle between them is π3\dfrac{\pi}{3}3π​. If λa⃗+2b⃗\lambda\vec{a}+2\vec{b}λa+2b and 3a⃗−λb⃗3\vec{a}-\lambda\vec{b}3a−λb are perpendicular to each other, then the number of values of λ\lambdaλ in [−1,3][-1,3][−1,3] is:
  1. (A)3
  2. (B)2
  3. (C)1
  4. (D)0

Correct answer: (D)

Step-by-step solution →
Q50·MathematicsIntegerJEE Main 2025
Let c^\hat cc^ be the projection vector of b⃗=λi^+4k^ (λ>0)\vec b=\lambda\hat i+4\hat k\ (\lambda>0)b=λi^+4k^ (λ>0) on a⃗=i^+2j^+2k^\vec a=\hat i+2\hat j+2\hat ka=i^+2j^​+2k^. If ∣a⃗+c^∣=7|\vec a+\hat c|=7∣a+c^∣=7, find the area of the parallelogram formed by b⃗\vec bb and c^\hat cc^.

Correct answer: 16

Step-by-step solution →
Q51·MathematicsIntegerJEE Advanced 2024
Let p⃗=2i^+j^+3k^\vec{p} = 2\hat{i} + \hat{j} + 3\hat{k}p​=2i^+j^​+3k^ and q⃗=i^−j^+k^\vec{q} = \hat{i} - \hat{j} + \hat{k}q​=i^−j^​+k^. If for some real numbers α\alphaα, β\betaβ, and γ\gammaγ, we have 15i^+10j^+6k^=α(2p⃗+q⃗)+β(p⃗−2q⃗)+γ(p⃗×q⃗)15\hat{i} + 10\hat{j} + 6\hat{k} = \alpha\left(2\vec{p} + \vec{q}\right) + \beta\left(\vec{p} - 2\vec{q}\right) + \gamma\left(\vec{p} \times \vec{q}\right)15i^+10j^​+6k^=α(2p​+q​)+β(p​−2q​)+γ(p​×q​), then the value of γ\gammaγ is ______

Correct answer: 2

Step-by-step solution →
Q52·MathematicsIntegerJEE Advanced 2024
Let OP→=α−1αi^+j^+k^,OQ→=i^+β−1βj^+k^\overrightarrow{OP} = \frac{\alpha - 1}{\alpha}\hat{i} + \hat{j} + \hat{k}, \overrightarrow{OQ} = \hat{i} + \frac{\beta - 1}{\beta}\hat{j} + \hat{k}OP=αα−1​i^+j^​+k^,OQ​=i^+ββ−1​j^​+k^ and OR→=i^+j^+12k^\overrightarrow{OR} = \hat{i} + \hat{j} + \frac{1}{2}\hat{k}OR=i^+j^​+21​k^ be three vectors, where α,β∈R−{0}\alpha, \beta \in R - \{0\}α,β∈R−{0} and OOO denotes the origin. If (OP→×OQ→)⋅OR→=0\left(\overrightarrow{OP} \times \overrightarrow{OQ}\right) \cdot \overrightarrow{OR} = 0(OP×OQ​)⋅OR=0 and the point (α,β,2)(\alpha, \beta, 2)(α,β,2) lies on the plane 3x+3y−z+l=03x + 3y - z + l = 03x+3y−z+l=0, then the value of lll is ______ .

Correct answer: 5

Step-by-step solution →
Q53·MathematicsSingle correctJEE Main 2024
Let three vectors a⃗=αi^+4j^+2k^\vec{a} = \alpha\hat{i} + 4\hat{j} + 2\hat{k}a=αi^+4j^​+2k^, b⃗=5i^+3j^+4k^\vec{b} = 5\hat{i} + 3\hat{j} + 4\hat{k}b=5i^+3j^​+4k^, c⃗=xi^+yj^+zk^\vec{c} = x\hat{i} + y\hat{j} + z\hat{k}c=xi^+yj^​+zk^ form a triangle such that c⃗=a⃗−b⃗\vec{c} = \vec{a} - \vec{b}c=a−b and the area of the triangle is 565\sqrt{6}56​. If α\alphaα is a positive real number, then ∣c⃗∣2|\vec{c}|^2∣c∣2 is :
  1. (A)16
  2. (B)14
  3. (C)12
  4. (D)10

Correct answer: (B)

Step-by-step solution →
Q54·MathematicsSingle correctJEE Main 2024
Between the following two statements: Statement-I: Let a⃗=i^+2j^−3k^\vec{a}=\hat{i}+2\hat{j}-3\hat{k}a=i^+2j^​−3k^ and b⃗=2i^+j^−k^\vec{b}=2\hat{i}+\hat{j}-\hat{k}b=2i^+j^​−k^. Then the vector r⃗\vec{r}r satisfying a⃗×r⃗=a⃗×b⃗\vec{a}\times\vec{r}=\vec{a}\times\vec{b}a×r=a×b and a⃗⋅r⃗=0\vec{a}\cdot\vec{r}=0a⋅r=0 is of magnitude 10\sqrt{10}10​. Statement-II: In a triangle ABC, cos⁡2A+cos⁡2B+cos⁡2C≥−32\cos 2A+\cos 2B+\cos 2C\ge -\dfrac{3}{2}cos2A+cos2B+cos2C≥−23​. In the light of the above statements, choose the correct answer:
  1. (A)Both Statement-I and Statement-II are incorrect
  2. (B)Statement-I is incorrect but Statement-II is correct
  3. (C)Both Statement-I and Statement-II are correct
  4. (D)Statement-I is correct but Statement-II is incorrect

Correct answer: (B)

Step-by-step solution →
Q55·MathematicsSingle correctJEE Main 2024
Let OA⃗=2a⃗\vec{OA} = 2\vec{a}OA=2a, OB⃗=6a⃗+5b⃗\vec{OB} = 6\vec{a} + 5\vec{b}OB=6a+5b and OC⃗=3b⃗\vec{OC} = 3\vec{b}OC=3b, where O is the origin. If the area of the parallelogram with adjacent sides OA⃗\vec{OA}OA and OC⃗\vec{OC}OC is 15 sq. units, then the area (in sq. units) of the quadrilateral OABC is equal to:
  1. (A)38
  2. (B)40
  3. (C)32
  4. (D)35

Correct answer: (D)

Step-by-step solution →
Q56·MathematicsSingle correctJEE Main 2024
Let a⃗=2i^+αj^+k^\vec{a}=2\hat{i}+\alpha\hat{j}+\hat{k}a=2i^+αj^​+k^, b⃗=−i^+k^\vec{b}=-\hat{i}+\hat{k}b=−i^+k^, c⃗=βj^−k^\vec{c}=\beta\hat{j}-\hat{k}c=βj^​−k^, where α\alphaα and β\betaβ are integers and αβ=−6\alpha\beta=-6αβ=−6. Let the values of the ordered pair (α,β)(\alpha, \beta)(α,β) for which the area of the parallelogram with diagonals a⃗+b⃗\vec{a}+\vec{b}a+b and b⃗+c⃗\vec{b}+\vec{c}b+c is 212\dfrac{\sqrt{21}}{2}221​​, be (α1,β1)(\alpha_{1}, \beta_{1})(α1​,β1​) and (α2,β2)(\alpha_{2}, \beta_{2})(α2​,β2​). Then α12+β12−α2β2\alpha_{1}^{2}+\beta_{1}^{2}-\alpha_{2}\beta_{2}α12​+β12​−α2​β2​ is equal to:
  1. (A)171717
  2. (B)242424
  3. (C)212121
  4. (D)191919

Correct answer: (D)

Step-by-step solution →
Q57·MathematicsNumericalJEE Main 2024
Let a⃗=9i^−13j^+25k^\vec a=9\hat i-13\hat j+25\hat ka=9i^−13j^​+25k^, b⃗=3i^+7j^−13k^\vec b=3\hat i+7\hat j-13\hat kb=3i^+7j^​−13k^ and c⃗=17i^−2j^+k^\vec c=17\hat i-2\hat j+\hat kc=17i^−2j^​+k^ be three given vectors. If r⃗\vec rr is a vector such that r⃗×a⃗=(b⃗+c⃗)×a⃗\vec r\times\vec a=(\vec b+\vec c)\times\vec ar×a=(b+c)×a and r⃗⋅(b⃗−c⃗)=0\vec r\cdot(\vec b-\vec c)=0r⋅(b−c)=0, then ∣593r⃗+67a⃗∣2(593)2\dfrac{\left|593\vec r+67\vec a\right|^2}{(593)^2}(593)2∣593r+67a∣2​ is equal to ___

Correct answer: 569

Step-by-step solution →
Q58·MathematicsSingle correctJEE Main 2024
The set of all α\alphaα, for which the vectors a⃗=αti^+6j^−3k^\vec a=\alpha t\hat i+6\hat j-3\hat ka=αti^+6j^​−3k^ and b⃗=ti^−2j^−2αtk^\vec b=t\hat i-2\hat j-2\alpha t\hat kb=ti^−2j^​−2αtk^ are inclined at an obtuse angle for all t∈Rt\in\mathbb{R}t∈R is:
  1. (A)[0,1)[0,1)[0,1)
  2. (B)(−2,0](-2,0](−2,0]
  3. (C)(−43,0]\left(-\dfrac43,0\right](−34​,0]
  4. (D)(−43,1]\left(-\dfrac43,1\right](−34​,1]

Correct answer: (C)

Step-by-step solution →
Q59·MathematicsSingle correctJEE Main 2024
Let a⃗=i^+2j^+3k^\vec{a}=\hat{i}+2\hat{j}+3\hat{k}a=i^+2j^​+3k^, b⃗=2i^+3j^−5k^\vec{b}=2\hat{i}+3\hat{j}-5\hat{k}b=2i^+3j^​−5k^ and c⃗=3i^−j^+λk^\vec{c}=3\hat{i}-\hat{j}+\lambda\hat{k}c=3i^−j^​+λk^ be three vectors. Let r⃗\vec{r}r be a unit vector along b⃗+c⃗\vec{b}+\vec{c}b+c. If r⃗⋅a⃗=3\vec{r}\cdot\vec{a}=3r⋅a=3, then 3λ3\lambda3λ is equal to
  1. (A)272727
  2. (B)252525
  3. (C)232323
  4. (D)212121

Correct answer: (B)

Step-by-step solution →
Q60·MathematicsSingle correctJEE Main 2024
Let a⃗=4i^−j^+k^\vec{a}=4\hat{i}-\hat{j}+\hat{k}a=4i^−j^​+k^, b⃗=11i^−j^+k^\vec{b}=11\hat{i}-\hat{j}+\hat{k}b=11i^−j^​+k^ and c⃗\vec{c}c be a vector such that (a⃗+b⃗)×c⃗=c⃗×(−2a⃗+3b⃗)(\vec{a}+\vec{b})\times\vec{c}=\vec{c}\times(-2\vec{a}+3\vec{b})(a+b)×c=c×(−2a+3b). If (2a⃗+3b⃗)⋅c⃗=1670(2\vec{a}+3\vec{b})\cdot\vec{c}=1670(2a+3b)⋅c=1670, then ∣c⃗∣2|\vec{c}|^2∣c∣2 is equal to
  1. (A)162716271627
  2. (B)161816181618
  3. (C)160016001600
  4. (D)160916091609

Correct answer: (B)

Step-by-step solution →
Q61·MathematicsSingle correctJEE Main 2024
If A(3,1,−1)A(3,1,-1)A(3,1,−1), B(53,73,13)B\left(\tfrac{5}{3},\tfrac{7}{3},\tfrac{1}{3}\right)B(35​,37​,31​), C(2,2,1)C(2,2,1)C(2,2,1) and D(103,23,−13)D\left(\tfrac{10}{3},\tfrac{2}{3},-\tfrac{1}{3}\right)D(310​,32​,−31​) are the vertices of a quadrilateral ABCDABCDABCD, then its area is
  1. (A)423\tfrac{4\sqrt{2}}{3}342​​
  2. (B)523\tfrac{5\sqrt{2}}{3}352​​
  3. (C)222\sqrt{2}22​
  4. (D)223\tfrac{2\sqrt{2}}{3}322​​

Correct answer: (A)

Step-by-step solution →
Q62·MathematicsSingle correctJEE Main 2024
Let a⃗=6i^+j^−k^\vec{a} = 6\hat{i} + \hat{j} - \hat{k}a=6i^+j^​−k^ and b⃗=i^+j^\vec{b} = \hat{i} + \hat{j}b=i^+j^​. If c⃗\vec{c}c is a vector such that ∣c⃗∣≥6|\vec{c}| \ge 6∣c∣≥6, a⃗⋅c⃗=6∣c⃗∣\vec{a}\cdot\vec{c} = 6|\vec{c}|a⋅c=6∣c∣, ∣c⃗−a⃗∣=22|\vec{c} - \vec{a}| = 2\sqrt{2}∣c−a∣=22​ and the angle between a⃗×b⃗\vec{a}\times\vec{b}a×b and c⃗\vec{c}c is 60∘60^\circ60∘, then ∣(a⃗×b⃗)×c⃗∣|(\vec{a}\times\vec{b})\times\vec{c}|∣(a×b)×c∣ is equal to:
  1. (A)92(6−6)\tfrac{9}{2}(6 - \sqrt{6})29​(6−6​)
  2. (B)323\tfrac{3}{2}\sqrt{3}23​3​
  3. (C)326\tfrac{3}{2}\sqrt{6}23​6​
  4. (D)92(6+6)\tfrac{9}{2}(6 + \sqrt{6})29​(6+6​)

Correct answer: (D)

Step-by-step solution →
Q63·MathematicsNumericalJEE Main 2024
Let a⃗=2i^−3j^+4k^\vec{a}=2\hat{i}-3\hat{j}+4\hat{k}a=2i^−3j^​+4k^, b⃗=3i^+4j^−5k^\vec{b}=3\hat{i}+4\hat{j}-5\hat{k}b=3i^+4j^​−5k^, and a vector c⃗\vec{c}c be such that a⃗×(b⃗+c⃗)+b⃗×c⃗=i^+8j^+13k^\vec{a}\times(\vec{b}+\vec{c})+\vec{b}\times\vec{c}=\hat{i}+8\hat{j}+13\hat{k}a×(b+c)+b×c=i^+8j^​+13k^. If a⃗⋅c⃗=13\vec{a}\cdot\vec{c}=13a⋅c=13, then 24−b⃗⋅c⃗24-\vec{b}\cdot\vec{c}24−b⋅c is equal to _______.

Correct answer: 46

Step-by-step solution →
Q64·MathematicsSingle correctJEE Main 2024
Let a⃗=2i^+j^−k^\vec{a}=2\hat{i}+\hat{j}-\hat{k}a=2i^+j^​−k^, b⃗=((a⃗×(i^+j^))×i^)×i^\vec{b}=\left(\left(\vec{a}\times(\hat{i}+\hat{j})\right)\times\hat{i}\right)\times\hat{i}b=((a×(i^+j^​))×i^)×i^. Then the square of the projection of a⃗\vec{a}a on b⃗\vec{b}b is:
  1. (A)15\frac{1}{5}51​
  2. (B)222
  3. (C)13\frac{1}{3}31​
  4. (D)23\frac{2}{3}32​

Correct answer: (B)

Step-by-step solution →
Q65·MathematicsSingle correctJEE Main 2024
Let a⃗=2i^+5j^−k^\vec{a}=2\hat{i}+5\hat{j}-\hat{k}a=2i^+5j^​−k^, b⃗=2i^−2j^+2k^\vec{b}=2\hat{i}-2\hat{j}+2\hat{k}b=2i^−2j^​+2k^ and c⃗\vec{c}c be three vectors such that (c⃗+i^)×(a⃗+b⃗+i^)=a⃗×(c⃗+i^)(\vec{c}+\hat{i})\times(\vec{a}+\vec{b}+\hat{i})=\vec{a}\times(\vec{c}+\hat{i})(c+i^)×(a+b+i^)=a×(c+i^) and a⃗⋅c⃗=−29\vec{a}\cdot\vec{c}=-29a⋅c=−29. Then c⃗⋅(−2i^+j^+k^)\vec{c}\cdot(-2\hat{i}+\hat{j}+\hat{k})c⋅(−2i^+j^​+k^) is equal to:
  1. (A)10
  2. (B)5
  3. (C)15
  4. (D)12

Correct answer: (B)

Step-by-step solution →
Q66·MathematicsNumericalJEE Main 2024
Let a⃗=i^−3j^+7k^\vec{a} = \hat{i} - 3\hat{j} + 7\hat{k}a=i^−3j^​+7k^, b⃗=2i^−j^+k^\vec{b} = 2\hat{i} - \hat{j} + \hat{k}b=2i^−j^​+k^ and c⃗\vec{c}c be a vector such that (a⃗+2b⃗)×c⃗=3(c⃗×a⃗)(\vec{a} + 2\vec{b}) \times \vec{c} = 3(\vec{c} \times \vec{a})(a+2b)×c=3(c×a). If a⃗⋅c⃗=130\vec{a} \cdot \vec{c} = 130a⋅c=130, then b⃗⋅c⃗\vec{b} \cdot \vec{c}b⋅c is equal to ___.

Correct answer: 30

Step-by-step solution →
Q67·MathematicsSingle correctJEE Main 2024
Consider three vectors a⃗,b⃗,c⃗\vec{a},\vec{b},\vec{c}a,b,c. Let ∣a⃗∣=2|\vec{a}|=2∣a∣=2, ∣b⃗∣=3|\vec{b}|=3∣b∣=3 and a⃗=b⃗×c⃗\vec{a}=\vec{b}\times\vec{c}a=b×c. If α∈[0,π3]\alpha\in\left[0,\dfrac{\pi}{3}\right]α∈[0,3π​] is the angle between the vectors b⃗\vec{b}b and c⃗\vec{c}c, then the minimum value of 27∣c⃗−a⃗∣227|\vec{c}-\vec{a}|^227∣c−a∣2 is equal to:
  1. (A)110
  2. (B)105
  3. (C)124
  4. (D)121

Correct answer: (C)

Step-by-step solution →
Q68·MathematicsSingle correctJEE Main 2024
If A(1,−1,2)A(1, -1, 2)A(1,−1,2), B(5,7,−6)B(5, 7, -6)B(5,7,−6), C(3,4,−10)C(3, 4, -10)C(3,4,−10) and D(−1,−4,−2)D(-1, -4, -2)D(−1,−4,−2) are the vertices of a quadrilateral ABCD, then its area is:
  1. (A)122912\sqrt{29}1229​
  2. (B)242924\sqrt{29}2429​
  3. (C)24724\sqrt{7}247​
  4. (D)48748\sqrt{7}487​

Correct answer: (A)

Step-by-step solution →
Q69·MathematicsSingle correctJEE Main 2024
Let a⃗=i^+j^+k^\vec{a}=\hat{i}+\hat{j}+\hat{k}a=i^+j^​+k^, b⃗=2i^+4j^−5k^\vec{b}=2\hat{i}+4\hat{j}-5\hat{k}b=2i^+4j^​−5k^ and c⃗=xi^+2j^+3k^\vec{c}=x\hat{i}+2\hat{j}+3\hat{k}c=xi^+2j^​+3k^, x∈Rx\in\mathbb{R}x∈R. If d⃗\vec{d}d is the unit vector in the direction of b⃗+c⃗\vec{b}+\vec{c}b+c such that a⃗⋅d⃗=1\vec{a}\cdot\vec{d}=1a⋅d=1, then (a⃗×b⃗)⋅c⃗(\vec{a}\times\vec{b})\cdot\vec{c}(a×b)⋅c is equal to
  1. (A)9
  2. (B)6
  3. (C)3
  4. (D)11

Correct answer: (D)

Step-by-step solution →
Q70·MathematicsSingle correctJEE Main 2024
Let a unit vector which makes an angle of 60∘60^\circ60∘ with 2i^+2j^−k^2\hat i+2\hat j-\hat k2i^+2j^​−k^ and an angle of 45∘45^\circ45∘ with i^−k^\hat i-\hat ki^−k^ be C⃗\vec CC. Then C⃗+(−12i^+132j^−23k^)\vec C+\left(-\dfrac12\hat i+\dfrac{1}{3\sqrt2}\hat j-\dfrac{\sqrt2}{3}\hat k\right)C+(−21​i^+32​1​j^​−32​​k^) is:
  1. (A)−23i^+23j^+(12+223)k^-\dfrac{\sqrt2}{3}\hat i+\dfrac{\sqrt2}{3}\hat j+\left(\dfrac12+\dfrac{2\sqrt2}{3}\right)\hat k−32​​i^+32​​j^​+(21​+322​​)k^
  2. (B)23i^+132j^−12k^\dfrac{\sqrt2}{3}\hat i+\dfrac{1}{3\sqrt2}\hat j-\dfrac12\hat k32​​i^+32​1​j^​−21​k^
  3. (C)(13+12)i^+(13−132)j^+(13+23)k^\left(\dfrac{1}{\sqrt3}+\dfrac12\right)\hat i+\left(\dfrac{1}{\sqrt3}-\dfrac{1}{3\sqrt2}\right)\hat j+\left(\dfrac{1}{\sqrt3}+\dfrac{\sqrt2}{3}\right)\hat k(3​1​+21​)i^+(3​1​−32​1​)j^​+(3​1​+32​​)k^
  4. (D)23i^−12k^\dfrac{\sqrt2}{3}\hat i-\dfrac12\hat k32​​i^−21​k^

Correct answer: (D)

Step-by-step solution →
Q71·MathematicsSingle correctJEE Main 2024
If λ>0\lambda>0λ>0, let θ\thetaθ be the angle between the vectors a⃗=i^+λj^−3k^\vec{a}=\hat{i}+\lambda\hat{j}-3\hat{k}a=i^+λj^​−3k^ and b⃗=3i^−j^+2k^\vec{b}=3\hat{i}-\hat{j}+2\hat{k}b=3i^−j^​+2k^. If the vectors a⃗+b⃗\vec{a}+\vec{b}a+b and a⃗−b⃗\vec{a}-\vec{b}a−b are mutually perpendicular, then the value of (14cos⁡θ)2(14\cos\theta)^2(14cosθ)2 is equal to
  1. (A)25
  2. (B)20
  3. (C)50
  4. (D)40

Correct answer: (A)

Step-by-step solution →
Q72·MathematicsNumericalJEE Main 2024
Let ABCABCABC be a triangle of area 15215\sqrt2152​ and the vectors AB⃗=i^+2j^−7k^\vec{AB}=\hat i+2\hat j-7\hat kAB=i^+2j^​−7k^, BC⃗=ai^+bj^+ck^\vec{BC}=a\hat i+b\hat j+c\hat kBC=ai^+bj^​+ck^ and AC⃗=6i^+dj^−2k^\vec{AC}=6\hat i+d\hat j-2\hat kAC=6i^+dj^​−2k^, d>0d>0d>0. Then the square of the length of the largest side of the triangle ABCABCABC is ___

Correct answer: 54

Step-by-step solution →
Q73·MathematicsNumericalJEE Main 2024
Let a⃗=i^+j^+k^\vec a=\hat i+\hat j+\hat ka=i^+j^​+k^, b⃗=−i^−8j^+2k^\vec b=-\hat i-8\hat j+2\hat kb=−i^−8j^​+2k^ and c⃗=4i^+c2j^+c3k^\vec c=4\hat i+c_2\hat j+c_3\hat kc=4i^+c2​j^​+c3​k^ be three vectors such that b⃗×a⃗=c⃗×a⃗\vec b\times\vec a=\vec c\times\vec ab×a=c×a. If the angle between the vector c⃗\vec cc and the vector 3i^+4j^+k^3\hat i+4\hat j+\hat k3i^+4j^​+k^ is θ\thetaθ, then the greatest integer less than or equal to tan⁡2θ\tan^2\thetatan2θ is __________.

Correct answer: 38

Step-by-step solution →
Q74·MathematicsSingle correctJEE Main 2024
Consider a △ABC\triangle ABC△ABC where A(1,2,3)A(1,2,3)A(1,2,3), B(−2,8,0)B(-2,8,0)B(−2,8,0) and C(3,6,7)C(3,6,7)C(3,6,7). If the angle bisector of ∠BAC\angle BAC∠BAC meets the line BC at D, then the length of the projection of the vector AD⃗\vec{AD}AD on the vector AC⃗\vec{AC}AC is:
  1. (A)37238\dfrac{37}{2\sqrt{38}}238​37​
  2. (B)382\dfrac{\sqrt{38}}{2}238​​
  3. (C)39238\dfrac{39}{2\sqrt{38}}238​39​
  4. (D)19\sqrt{19}19​

Correct answer: (A)

Step-by-step solution →
Q75·MathematicsSingle correctJEE Main 2024
Let a⃗=−5i^+j^−3k^\vec a=-5\hat i+\hat j-3\hat ka=−5i^+j^​−3k^, b⃗=i^+2j^−4k^\vec b=\hat i+2\hat j-4\hat kb=i^+2j^​−4k^ and c⃗=((((a⃗×b⃗)×i^)×i^)×i^)\vec c=\big(\big(\big((\vec a\times\vec b)\times\hat i\big)\times\hat i\big)\times\hat i\big)c=((((a×b)×i^)×i^)×i^). Then c⃗⋅(−i^+j^+k^)\vec c\cdot(-\hat i+\hat j+\hat k)c⋅(−i^+j^​+k^) is equal to:
  1. (A)−12-12−12
  2. (B)−10-10−10
  3. (C)−13-13−13
  4. (D)−15-15−15

Correct answer: (A)

Step-by-step solution →
Q76·MathematicsSingle correctJEE Main 2024
Let a⃗=3i^+j^−2k^\vec{a}=3\hat{i}+\hat{j}-2\hat{k}a=3i^+j^​−2k^, b⃗=4i^+j^+7k^\vec{b}=4\hat{i}+\hat{j}+7\hat{k}b=4i^+j^​+7k^ and c⃗=i^−3j^+4k^\vec{c}=\hat{i}-3\hat{j}+4\hat{k}c=i^−3j^​+4k^ be three vectors. If a vector p⃗\vec{p}p​ satisfies p⃗×b⃗=c⃗×b⃗\vec{p}\times\vec{b}=\vec{c}\times\vec{b}p​×b=c×b and p⃗⋅a⃗=0\vec{p}\cdot\vec{a}=0p​⋅a=0, then p⃗⋅(i^−j^−k^)\vec{p}\cdot\left(\hat{i}-\hat{j}-\hat{k}\right)p​⋅(i^−j^​−k^) is equal to
  1. (A)242424
  2. (B)363636
  3. (C)282828
  4. (D)323232

Correct answer: (D)

Step-by-step solution →
Q77·MathematicsNumericalJEE Main 2024
Let a⃗\vec{a}a and b⃗\vec{b}b be two vectors such that ∣a⃗∣=1|\vec{a}|=1∣a∣=1, ∣b⃗∣=4|\vec{b}|=4∣b∣=4 and a⃗⋅b⃗=2\vec{a}\cdot\vec{b}=2a⋅b=2. If c⃗=(2a⃗×b⃗)−3b⃗\vec{c}=(2\vec{a}\times\vec{b})-3\vec{b}c=(2a×b)−3b and the angle between b⃗\vec{b}b and c⃗\vec{c}c is α\alphaα, then 192sin⁡2α192\sin^2\alpha192sin2α is equal to ______.

Correct answer: 48

Step-by-step solution →
Q78·MathematicsNumericalJEE Main 2024
Let a⃗=3i^+2j^+k^\vec{a}=3\hat{i}+2\hat{j}+\hat{k}a=3i^+2j^​+k^, b⃗=2i^−j^+3k^\vec{b}=2\hat{i}-\hat{j}+3\hat{k}b=2i^−j^​+3k^ and c⃗\vec{c}c be a vector such that (a⃗+b⃗)×c⃗=2(a⃗×b⃗)+24j^−6k^(\vec{a}+\vec{b})\times\vec{c}=2(\vec{a}\times\vec{b})+24\hat{j}-6\hat{k}(a+b)×c=2(a×b)+24j^​−6k^ and (a⃗−b⃗+i^)⋅c⃗=−3(\vec{a}-\vec{b}+\hat{i})\cdot\vec{c}=-3(a−b+i^)⋅c=−3. Then ∣c⃗∣2|\vec{c}|^2∣c∣2 is equal to ______.

Correct answer: 38

Step-by-step solution →
Q79·MathematicsSingle correctJEE Main 2024
Let a⃗=i^+αj^+βk^\vec{a}=\hat{i}+\alpha\hat{j}+\beta\hat{k}a=i^+αj^​+βk^, α,β∈R\alpha,\beta\in Rα,β∈R. Let a vector b⃗\vec{b}b be such that the angle between a⃗\vec{a}a and b⃗\vec{b}b is π4\dfrac{\pi}{4}4π​ and ∣b⃗∣2=6|\vec{b}|^2=6∣b∣2=6. If a⃗⋅b⃗=32\vec{a}\cdot\vec{b}=3\sqrt{2}a⋅b=32​, then the value of (α2+β2) ∣a⃗×b⃗∣2(\alpha^2+\beta^2)\,|\vec{a}\times\vec{b}|^2(α2+β2)∣a×b∣2 is equal to:
  1. (A)909090
  2. (B)757575
  3. (C)959595
  4. (D)858585

Correct answer: (A)

Step-by-step solution →
Q80·MathematicsSingle correctJEE Main 2024
Let A(2,3,5)A(2,3,5)A(2,3,5) and C(−3,4,−2)C(-3,4,-2)C(−3,4,−2) be opposite vertices of a parallelogram ABCDABCDABCD. If the diagonal BD⃗=i^+2j^+3k^\vec{BD}=\hat i+2\hat j+3\hat kBD=i^+2j^​+3k^, then the area of the parallelogram is equal to:
  1. (A)12410\dfrac12\sqrt{410}21​410​
  2. (B)12474\dfrac12\sqrt{474}21​474​
  3. (C)12586\dfrac12\sqrt{586}21​586​
  4. (D)12306\dfrac12\sqrt{306}21​306​

Correct answer: (B)

Step-by-step solution →
Q81·MathematicsSingle correctJEE Main 2024
Let a⃗\vec{a}a and b⃗\vec{b}b be two vectors such that ∣b⃗∣=1|\vec{b}|=1∣b∣=1 and ∣b⃗×a⃗∣=2|\vec{b}\times\vec{a}|=2∣b×a∣=2. Then ∣(b⃗×a⃗)−b⃗∣2\left|(\vec{b}\times\vec{a})-\vec{b}\right|^2​(b×a)−b​2 is equal to:
  1. (A)333
  2. (B)555
  3. (C)111
  4. (D)444

Correct answer: (B)

Step-by-step solution →
Q82·MathematicsSingle correctJEE Main 2024
Let a⃗=a1i^+a2j^+a3k^\vec a=a_1\hat i+a_2\hat j+a_3\hat ka=a1​i^+a2​j^​+a3​k^ and b⃗=b1i^+b2j^+b3k^\vec b=b_1\hat i+b_2\hat j+b_3\hat kb=b1​i^+b2​j^​+b3​k^ be two vectors such that ∣a⃗∣=1|\vec a|=1∣a∣=1, a⃗⋅b⃗=2\vec a\cdot\vec b=2a⋅b=2 and ∣b⃗∣=4|\vec b|=4∣b∣=4. If c⃗=2(a⃗×b⃗)−3b⃗\vec c=2(\vec a\times\vec b)-3\vec bc=2(a×b)−3b, then the angle between b⃗\vec bb and c⃗\vec cc is equal to:
  1. (A)cos⁡−1(23)\cos^{-1}\left(\dfrac{2}{\sqrt3}\right)cos−1(3​2​)
  2. (B)cos⁡−1(−13)\cos^{-1}\left(-\dfrac{1}{\sqrt3}\right)cos−1(−3​1​)
  3. (C)cos⁡−1(−32)\cos^{-1}\left(-\dfrac{\sqrt3}{2}\right)cos−1(−23​​)
  4. (D)cos⁡−1(23)\cos^{-1}\left(\dfrac{2}{3}\right)cos−1(32​)

Correct answer: (C)

Step-by-step solution →
Q83·MathematicsSingle correctJEE Main 2024
Let a unit vector u^=xi^+yj^+zk^\hat u=x\hat i+y\hat j+z\hat ku^=xi^+yj^​+zk^ make angles π2,π3\dfrac{\pi}{2},\dfrac{\pi}{3}2π​,3π​ and 2π3\dfrac{2\pi}{3}32π​ with the vectors 12i^+12j^\dfrac{1}{\sqrt2}\hat i+\dfrac{1}{\sqrt2}\hat j2​1​i^+2​1​j^​, 12j^+12k^\dfrac{1}{\sqrt2}\hat j+\dfrac{1}{\sqrt2}\hat k2​1​j^​+2​1​k^ and 12i^+12k^\dfrac{1}{\sqrt2}\hat i+\dfrac{1}{\sqrt2}\hat k2​1​i^+2​1​k^ respectively. If v⃗=12i^+12j^+12k^\vec v=\dfrac{1}{\sqrt2}\hat i+\dfrac{1}{\sqrt2}\hat j+\dfrac{1}{\sqrt2}\hat kv=2​1​i^+2​1​j^​+2​1​k^, then ∣u^−v⃗∣2|\hat u-\vec v|^2∣u^−v∣2 is equal to:
  1. (A)112\dfrac{11}{2}211​
  2. (B)52\dfrac{5}{2}25​
  3. (C)9
  4. (D)7

Correct answer: (B)

Step-by-step solution →
Q84·MathematicsSingle correctJEE Main 2024
Let O be the origin and the position vector of A and B be 2i^+2j^+k^2\hat{i}+2\hat{j}+\hat{k}2i^+2j^​+k^ and 2i^+4j^+4k^2\hat{i}+4\hat{j}+4\hat{k}2i^+4j^​+4k^ respectively. If the internal bisector of ∠AOB\angle AOB∠AOB meets the line AB at C, then the length of OC is
  1. (A)2331\dfrac{2}{3}\sqrt{31}32​31​
  2. (B)2334\dfrac{2}{3}\sqrt{34}32​34​
  3. (C)3434\dfrac{3}{4}\sqrt{34}43​34​
  4. (D)3231\dfrac{3}{2}\sqrt{31}23​31​

Correct answer: (B)

Step-by-step solution →
Q85·MathematicsSingle correctJEE Main 2024
Let OA→=a⃗, OB→=12a⃗+4b⃗\overrightarrow{OA}=\vec a,\ \overrightarrow{OB}=12\vec a+4\vec bOA=a, OB=12a+4b and OC→=b⃗\overrightarrow{OC}=\vec bOC=b, where OOO is the origin. If SSS is the parallelogram with adjacent sides OAOAOA and OCOCOC, then area of the quadrilateral OABCarea of S\dfrac{\text{area of the quadrilateral }OABC}{\text{area of }S}area of Sarea of the quadrilateral OABC​ is equal to:
  1. (A)6
  2. (B)10
  3. (C)7
  4. (D)8

Correct answer: (D)

Step-by-step solution →
Q86·MathematicsSingle correctJEE Main 2024
Let a⃗,b⃗\vec{a},\vec{b}a,b and c⃗\vec{c}c be three non-zero vectors such that b⃗\vec{b}b and c⃗\vec{c}c are non-collinear. If a⃗+5b⃗\vec{a}+5\vec{b}a+5b is collinear with c⃗\vec{c}c, b⃗+6c⃗\vec{b}+6\vec{c}b+6c is collinear with a⃗\vec{a}a and a⃗+αb⃗+βc⃗=0⃗\vec{a}+\alpha\vec{b}+\beta\vec{c}=\vec{0}a+αb+βc=0, then α+β\alpha+\betaα+β is equal to
  1. (A)353535
  2. (B)303030
  3. (C)−30-30−30
  4. (D)−25-25−25

Correct answer: (A)

Step-by-step solution →
Q87·MathematicsSingle correctJEE Main 2024
Let the position vectors of the vertices A, B and C of a triangle be 2i^+2j^+k^2\hat i+2\hat j+\hat k2i^+2j^​+k^, i^+2j^+2k^\hat i+2\hat j+2\hat ki^+2j^​+2k^ and 2i^+j^+2k^2\hat i+\hat j+2\hat k2i^+j^​+2k^ respectively. Let l1l_1l1​, l2l_2l2​ and l3l_3l3​ be the lengths of perpendiculars drawn from the ortho center of the triangle on the sides AB, BC and CA respectively, then l12+l22+l32l_1^2+l_2^2+l_3^2l12​+l22​+l32​ equals :
  1. (A)15\dfrac1551​
  2. (B)12\dfrac1221​
  3. (C)14\dfrac1441​
  4. (D)13\dfrac1331​

Correct answer: (B)

Step-by-step solution →
Q88·MathematicsSingle correctJEE Main 2024
Let a⃗=i^+2j^+k^\vec a=\hat i+2\hat j+\hat ka=i^+2j^​+k^, b⃗=3(i^−j^+k^)\vec b=3(\hat i-\hat j+\hat k)b=3(i^−j^​+k^). Let c⃗\vec cc be the vector such that a⃗×c⃗=b⃗\vec a\times\vec c=\vec ba×c=b and a⃗⋅c⃗=3\vec a\cdot\vec c=3a⋅c=3. Then a⃗⋅((c⃗×b⃗)−b⃗−c⃗)\vec a\cdot\big((\vec c\times\vec b)-\vec b-\vec c\big)a⋅((c×b)−b−c) is equal to:
  1. (A)32
  2. (B)24
  3. (C)20
  4. (D)36

Correct answer: (B)

Step-by-step solution →
Q89·MathematicsSingle correctJEE Main 2024
The position vectors of the vertices A, B and C of a triangle are 2i^−3j^+3k^2\hat i-3\hat j+3\hat k2i^−3j^​+3k^, 2i^+2j^+3k^2\hat i+2\hat j+3\hat k2i^+2j^​+3k^ and −i^+j^+3k^-\hat i+\hat j+3\hat k−i^+j^​+3k^ respectively. Let lll denotes the length of the angle bisector AD of ∠BAC\angle BAC∠BAC where D is on the line segment BC, then 2l22l^22l2 equals :
  1. (A)49
  2. (B)42
  3. (C)50
  4. (D)45

Correct answer: (D)

Step-by-step solution →
Q90·MathematicsNumericalJEE Main 2024
The least positive integral value of α\alphaα, for which the angle between the vectors αi^−2j^+2k^\alpha\hat i-2\hat j+2\hat kαi^−2j^​+2k^ and αi^+2αj^−2k^\alpha\hat i+2\alpha\hat j-2\hat kαi^+2αj^​−2k^ is acute, is __________.

Correct answer: 5

Step-by-step solution →
Q91·MathematicsNumericalJEE Advanced 2023
Let P be the plane 3x+2y+3z=16\sqrt{3}x + 2y + 3z = 163​x+2y+3z=16 and let S={αi^+βj^+γk^:α2+β2+γ2=1 and the distance of (α,β,γ) from the plane P is 72}S = \left\{\alpha\hat{i} + \beta\hat{j} + \gamma\hat{k} : \alpha^2 + \beta^2 + \gamma^2 = 1 \text{ and the distance of } (\alpha, \beta, \gamma) \text{ from the plane P is } \frac{7}{2}\right\}S={αi^+βj^​+γk^:α2+β2+γ2=1 and the distance of (α,β,γ) from the plane P is 27​}. Let u⃗,v⃗\vec{u}, \vec{v}u,v and w⃗\vec{w}w be three distinct vectors in S such that ∣u⃗−v⃗∣=∣v⃗−w⃗∣=∣w⃗−u⃗∣\left|\vec{u} - \vec{v}\right| = \left|\vec{v} - \vec{w}\right| = \left|\vec{w} - \vec{u}\right|∣u−v∣=∣v−w∣=∣w−u∣. Let V be the volume of the parallelepiped determined by vectors, u⃗,v⃗\vec{u}, \vec{v}u,v and w⃗\vec{w}w . Then the value of 803V\frac{80}{\sqrt{3}}V3​80​V is

Correct answer: 45

Step-by-step solution →
Q92·MathematicsSingle correctJEE Advanced 2023
Let the position vectors of the points PPP, QQQ, RRR and S be a⃗=i^+2j^−5k^\vec{a} = \hat{i} + 2\hat{j} - 5\hat{k}a=i^+2j^​−5k^, b⃗=3i^+6j^+3k^\vec{b} = 3\hat{i} + 6\hat{j} + 3\hat{k}b=3i^+6j^​+3k^, c⃗=175i^+165j^+7k^\vec{c} = \frac{17}{5}\hat{i} + \frac{16}{5}\hat{j} + 7\hat{k}c=517​i^+516​j^​+7k^ and d⃗=2i^+j^+k^\vec{d} = 2\hat{i} + \hat{j} + \hat{k}d=2i^+j^​+k^, respectively. Then which of the following statements is true?
  1. (A)The points PPP, QQQ, RRR and S are NOT coplanar
  2. (B)b⃗+2d⃗3\frac{\vec{b} + 2\vec{d}}{3}3b+2d​ is the position vector of a point which divides PRPRPR internally in the ratio 5:45 : 45:4
  3. (C)b⃗+2d⃗3\frac{\vec{b} + 2\vec{d}}{3}3b+2d​ is the position vector of a point which divides PRPRPR externally in the ratio 5:45 : 45:4
  4. (D)The square of the magnitude of the vector b⃗×d⃗\vec{b} \times \vec{d}b×d is 95

Correct answer: (B)

Step-by-step solution →
Q93·MathematicsSingle correctJEE Main 2023
Let SSS be the set of all (λ,μ)(\lambda, \mu)(λ,μ) for which the vectors λi^−j^+k^\lambda\hat{i} - \hat{j} + \hat{k}λi^−j^​+k^, i^+2j^+μk^\hat{i} + 2\hat{j} + \mu\hat{k}i^+2j^​+μk^ and 3i^−4j^+5k^3\hat{i} - 4\hat{j} + 5\hat{k}3i^−4j^​+5k^, where λ−μ=5\lambda - \mu = 5λ−μ=5, are coplanar, then ∑(λ,μ)∈S80(λ2+μ2)\sum_{(\lambda,\mu)\in S} 80(\lambda^2 + \mu^2)∑(λ,μ)∈S​80(λ2+μ2) is equal to:
  1. (A)237023702370
  2. (B)213021302130
  3. (C)229022902290
  4. (D)221022102210

Correct answer: (C)

Step-by-step solution →
Q94·MathematicsSingle correctJEE Main 2023
Let ABCD be a quadrilateral. If E and F are the mid points of the diagonals AC and BD respectively and (AB⃗−BC⃗)+(AD⃗−DC⃗)=k FE⃗\left(\vec{AB}-\vec{BC}\right)+\left(\vec{AD}-\vec{DC}\right)=k\,\vec{FE}(AB−BC)+(AD−DC)=kFE, then k is equal to
  1. (A)2
  2. (B)−2-2−2
  3. (C)−4-4−4
  4. (D)4

Correct answer: (C)

Step-by-step solution →
Q95·MathematicsSingle correctJEE Main 2023
Let ∣a⃗∣=2|\vec{a}| = 2∣a∣=2, ∣b⃗∣=3|\vec{b}| = 3∣b∣=3 and the angle between the vectors a⃗\vec{a}a and b⃗\vec{b}b be π4\dfrac{\pi}{4}4π​. Then ∣(a⃗+2b⃗)×(2a⃗−3b⃗)∣2\left|(\vec{a} + 2\vec{b}) \times (2\vec{a} - 3\vec{b})\right|^2​(a+2b)×(2a−3b)​2 is equal to
  1. (A)482
  2. (B)441
  3. (C)841
  4. (D)882

Correct answer: (D)

Step-by-step solution →
Q96·MathematicsSingle correctJEE Main 2023
Let a⃗=i^+4j^+2k^, b⃗=3i^−2j^+7k^\vec a=\hat i+4\hat j+2\hat k,\,\vec b=3\hat i-2\hat j+7\hat ka=i^+4j^​+2k^,b=3i^−2j^​+7k^ and c⃗=2i^−j^+4k^\vec c=2\hat i-\hat j+4\hat kc=2i^−j^​+4k^. If a vector d⃗\vec dd satisfies d⃗×b⃗=c⃗×b⃗\vec d\times\vec b=\vec c\times\vec bd×b=c×b and a⃗⋅d⃗=24\vec a\cdot\vec d=24a⋅d=24, then ∣d⃗∣2|\vec d|^{2}∣d∣2 is equal to:
  1. (A)413413413
  2. (B)423423423
  3. (C)323323323
  4. (D)313313313

Correct answer: (A)

Step-by-step solution →
Q97·MathematicsSingle correctJEE Main 2023
Let for a triangle ABC, AB→=−2i^+j^+3k^\overrightarrow{AB} = -2\hat{i} + \hat{j} + 3\hat{k}AB=−2i^+j^​+3k^, CB→=αi^+βj^+γk^\overrightarrow{CB} = \alpha\hat{i} + \beta\hat{j} + \gamma\hat{k}CB=αi^+βj^​+γk^, CA→=4i^+3j^+δk^\overrightarrow{CA} = 4\hat{i} + 3\hat{j} + \delta\hat{k}CA=4i^+3j^​+δk^. If δ>0\delta > 0δ>0 and the area of the triangle ABC is 565\sqrt{6}56​, then CB→⋅CA→\overrightarrow{CB} \cdot \overrightarrow{CA}CB⋅CA is equal to
  1. (A)60
  2. (B)120
  3. (C)108
  4. (D)54

Correct answer: (A)

Step-by-step solution →
Q98·MathematicsNumericalJEE Main 2023
Let a⃗=3i^+j^−k^\vec a=3\hat i+\hat j-\hat ka=3i^+j^​−k^ and c⃗=2i^−3j^+3k^\vec c=2\hat i-3\hat j+3\hat kc=2i^−3j^​+3k^. If b⃗\vec bb is a vector such that a⃗=b⃗×c⃗\vec a=\vec b\times\vec ca=b×c and ∣b⃗∣=50|\vec b|=\sqrt{50}∣b∣=50​, then ∣72−∣b⃗+c⃗∣2∣\left|72-|\vec b+\vec c|^{2}\right|​72−∣b+c∣2​ is equal to _____.

Correct answer: 66

Step-by-step solution →
Q99·MathematicsSingle correctJEE Main 2023
Let λ∈Z\lambda\in\mathbb{Z}λ∈Z, a⃗=2λi^+j^−k^\vec a=2\lambda\hat i+\hat j-\hat ka=2λi^+j^​−k^ and b⃗=3i^−j^+2k^\vec b=3\hat i-\hat j+2\hat kb=3i^−j^​+2k^. Let c⃗\vec cc be a vector such that (a⃗+b⃗+c⃗)×c⃗=0⃗(\vec a+\vec b+\vec c)\times\vec c=\vec 0(a+b+c)×c=0, a⃗⋅c⃗=−17\vec a\cdot\vec c=-17a⋅c=−17 and b⃗⋅c⃗=−20\vec b\cdot\vec c=-20b⋅c=−20. Then ∣c⃗×(λi^+j^+k^)∣2|\vec c\times(\lambda\hat i+\hat j+\hat k)|^2∣c×(λi^+j^​+k^)∣2 is equal to
  1. (A)62
  2. (B)46
  3. (C)53
  4. (D)49

Correct answer: (B)

Step-by-step solution →
Q100·MathematicsSingle correctJEE Main 2023
Let a, b, c be three distinct real numbers, none equal to one. If the vectors ai^+j^+k^a\hat i+\hat j+\hat kai^+j^​+k^, i^+bj^+k^\hat i+b\hat j+\hat ki^+bj^​+k^ and i^+j^+ck^\hat i+\hat j+c\hat ki^+j^​+ck^ are coplanar, then 11−a+11−b+11−c\dfrac{1}{1-a}+\dfrac{1}{1-b}+\dfrac{1}{1-c}1−a1​+1−b1​+1−c1​ is equal to
  1. (A)1
  2. (B)-1
  3. (C)3
  4. (D)-2

Correct answer: (A)

Step-by-step solution →
Q101·MathematicsSingle correctJEE Main 2023
If four distinct points with position vectors a⃗,b⃗,c⃗\vec{a},\vec{b},\vec{c}a,b,c and d⃗\vec{d}d are coplanar, then [a⃗ b⃗ c⃗]\left[\vec{a}\,\vec{b}\,\vec{c}\right][abc] is equal to
  1. (A)[d⃗ c⃗ a⃗]+[b⃗ a⃗ d⃗]+[c⃗ d⃗ b⃗]\left[\vec{d}\,\vec{c}\,\vec{a}\right]+\left[\vec{b}\,\vec{a}\,\vec{d}\right]+\left[\vec{c}\,\vec{d}\,\vec{b}\right][dca]+[bad]+[cdb]
  2. (B)[d⃗ b⃗ a⃗]+[a⃗ c⃗ d⃗]+[d⃗ b⃗ c⃗]\left[\vec{d}\,\vec{b}\,\vec{a}\right]+\left[\vec{a}\,\vec{c}\,\vec{d}\right]+\left[\vec{d}\,\vec{b}\,\vec{c}\right][dba]+[acd]+[dbc]
  3. (C)[a⃗ d⃗ b⃗]+[d⃗ c⃗ a⃗]+[b⃗ d⃗ c⃗]\left[\vec{a}\,\vec{d}\,\vec{b}\right]+\left[\vec{d}\,\vec{c}\,\vec{a}\right]+\left[\vec{b}\,\vec{d}\,\vec{c}\right][adb]+[dca]+[bdc]
  4. (D)[b⃗ c⃗ d⃗]+[a⃗ b⃗ c⃗]+[d⃗ b⃗ a⃗]\left[\vec{b}\,\vec{c}\,\vec{d}\right]+\left[\vec{a}\,\vec{b}\,\vec{c}\right]+\left[\vec{d}\,\vec{b}\,\vec{a}\right][bcd]+[abc]+[dba]

Correct answer: (A)

Step-by-step solution →
Q102·MathematicsSingle correctJEE Main 2023
For any vector a⃗=a1i^+a2j^+a3k^\vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}a=a1​i^+a2​j^​+a3​k^, with ∣ai∣<1|a_i| < 1∣ai​∣<1, i=1,2,3i = 1, 2, 3i=1,2,3, consider the following statements: (A): max⁡{∣a1∣,∣a2∣,∣a3∣}≤∣a⃗∣\max \{ |a_1|, |a_2|, |a_3| \} \le |\vec{a}|max{∣a1​∣,∣a2​∣,∣a3​∣}≤∣a∣ (B): ∣a⃗∣≤3max⁡{∣a1∣,∣a2∣,∣a3∣}|\vec{a}| \le 3 \max \{ |a_1|, |a_2|, |a_3| \}∣a∣≤3max{∣a1​∣,∣a2​∣,∣a3​∣}
  1. (A)Only (B) is true
  2. (B)Only (A) is true
  3. (C)Neither (A) nor (B) is true
  4. (D)Both (A) and (B) are true

Correct answer: (D)

Step-by-step solution →
Q103·MathematicsNumericalJEE Main 2023
Let a⃗=i^+2j^+3k^\vec{a}=\hat{i}+2\hat{j}+3\hat{k}a=i^+2j^​+3k^ and b⃗=i^+j^−k^\vec{b}=\hat{i}+\hat{j}-\hat{k}b=i^+j^​−k^. If c⃗\vec{c}c is a vector such that a⃗⋅c⃗=11\vec{a}\cdot\vec{c}=11a⋅c=11, b⃗⋅(a⃗×c⃗)=27\vec{b}\cdot(\vec{a}\times\vec{c})=27b⋅(a×c)=27 and b⃗⋅c⃗=−3 ∣b⃗∣\vec{b}\cdot\vec{c}=-\sqrt{3}\,|\vec{b}|b⋅c=−3​∣b∣, then ∣a⃗×c⃗∣2|\vec{a}\times\vec{c}|^2∣a×c∣2 is equal to

Correct answer: 285

Step-by-step solution →
Q104·MathematicsSingle correctJEE Main 2023
An arc PQ of a circle subtends a right angle at its centre O. The mid point of the arc PQ is R. If OP→=u⃗\overrightarrow{OP}=\vec uOP=u, OR→=v⃗\overrightarrow{OR}=\vec vOR=v and OQ→=αu⃗+βv⃗\overrightarrow{OQ}=\alpha\vec u+\beta\vec vOQ​=αu+βv, then α,β2\alpha,\beta^2α,β2 are the roots of the equation
  1. (A)x2−x−2=0x^2-x-2=0x2−x−2=0
  2. (B)3x2+2x−1=03x^2+2x-1=03x2+2x−1=0
  3. (C)x2+x−2=0x^2+x-2=0x2+x−2=0
  4. (D)3x2−2x−1=03x^2-2x-1=03x2−2x−1=0

Correct answer: (A)

Step-by-step solution →
Q105·MathematicsSingle correctJEE Main 2023
If the points P and Q are respectively the circumcentre and the orthocentre of a △ABC\triangle ABC△ABC, then PA→+PB→+PC→\overrightarrow{PA}+\overrightarrow{PB}+\overrightarrow{PC}PA+PB+PC is equal to:
  1. (A)2QP→2\overrightarrow{QP}2QP​
  2. (B)QP→\overrightarrow{QP}QP​
  3. (C)2PQ→2\overrightarrow{PQ}2PQ​
  4. (D)PQ→\overrightarrow{PQ}PQ​

Correct answer: (D)

Step-by-step solution →
Q106·MathematicsSingle correctJEE Main 2023
Let a⃗=2i^+7j^−k^\vec{a}=2\hat{i}+7\hat{j}-\hat{k}a=2i^+7j^​−k^, b⃗=3i^+5k^\vec{b}=3\hat{i}+5\hat{k}b=3i^+5k^ and c⃗=i^−j^+2k^\vec{c}=\hat{i}-\hat{j}+2\hat{k}c=i^−j^​+2k^. Let d⃗\vec{d}d be a vector which is perpendicular to both a⃗\vec{a}a and b⃗\vec{b}b, and c⃗⋅d⃗=12\vec{c}\cdot\vec{d}=12c⋅d=12. Then (−i^+j^−k^)⋅(c⃗×d⃗)\left(-\hat{i}+\hat{j}-\hat{k}\right)\cdot\left(\vec{c}\times\vec{d}\right)(−i^+j^​−k^)⋅(c×d) is equal to:
  1. (A)484848
  2. (B)424242
  3. (C)444444
  4. (D)242424

Correct answer: (C)

Step-by-step solution →
Q107·MathematicsSingle correctJEE Main 2023
Let O be the origin and the position vector of the point P be −i^−2j^+3k^-\hat i-2\hat j+3\hat k−i^−2j^​+3k^. If the position vectors of the points A, B and C are −2i^+j^−3k^-2\hat i+\hat j-3\hat k−2i^+j^​−3k^, 2i^+4j^−2k^2\hat i+4\hat j-2\hat k2i^+4j^​−2k^ and −4i^+2j^−k^-4\hat i+2\hat j-\hat k−4i^+2j^​−k^ respectively, then the projection of the vector OP→\overrightarrow{OP}OP on a vector perpendicular to the vectors AB→\overrightarrow{AB}AB and AC→\overrightarrow{AC}AC is
  1. (A)333
  2. (B)83\dfrac8338​
  3. (C)103\dfrac{10}{3}310​
  4. (D)73\dfrac7337​

Correct answer: (A)

Step-by-step solution →
Q108·MathematicsNumericalJEE Main 2023
Let a⃗=6i^+9j^+12k^\vec{a}=6\hat{i}+9\hat{j}+12\hat{k}a=6i^+9j^​+12k^, b⃗=αi^+11j^−2k^\vec{b}=\alpha\hat{i}+11\hat{j}-2\hat{k}b=αi^+11j^​−2k^ and c⃗\vec{c}c be vectors such that a⃗×c⃗=a⃗×b⃗\vec{a}\times\vec{c}=\vec{a}\times\vec{b}a×c=a×b. If a⃗⋅c⃗=−12\vec{a}\cdot\vec{c}=-12a⋅c=−12, c⃗⋅(i^−2j^+k^)=5\vec{c}\cdot(\hat{i}-2\hat{j}+\hat{k})=5c⋅(i^−2j^​+k^)=5, then c⃗⋅(i^+j^+k^)\vec{c}\cdot(\hat{i}+\hat{j}+\hat{k})c⋅(i^+j^​+k^) is equal to

Correct answer: 11

Step-by-step solution →
Q109·MathematicsSingle correctJEE Main 2023
Let the vectors u⃗1=i^+j^+ak^\vec{u}_1=\hat{i}+\hat{j}+a\hat{k}u1​=i^+j^​+ak^, u⃗2=i^+bj^+k^\vec{u}_2=\hat{i}+b\hat{j}+\hat{k}u2​=i^+bj^​+k^ and u⃗3=ci^+j^+k^\vec{u}_3=c\hat{i}+\hat{j}+\hat{k}u3​=ci^+j^​+k^ be coplanar. If the vectors v⃗1=(a+b)i^+cj^+ck^\vec{v}_1=(a+b)\hat{i}+c\hat{j}+c\hat{k}v1​=(a+b)i^+cj^​+ck^, v⃗2=ai^+(b+c)j^+ak^\vec{v}_2=a\hat{i}+(b+c)\hat{j}+a\hat{k}v2​=ai^+(b+c)j^​+ak^ and v⃗3=bi^+bj^+(c+a)k^\vec{v}_3=b\hat{i}+b\hat{j}+(c+a)\hat{k}v3​=bi^+bj^​+(c+a)k^ are also coplanar, then 6(a+b+c)6(a+b+c)6(a+b+c) is equal to
  1. (A)0
  2. (B)6
  3. (C)12
  4. (D)4

Correct answer: (C)

Step-by-step solution →
Q110·MathematicsSingle correctJEE Main 2023
If the points with vectors αi^+10j^+13k^\alpha\hat{i}+10\hat{j}+13\hat{k}αi^+10j^​+13k^, 6i^+11j^+11k^6\hat{i}+11\hat{j}+11\hat{k}6i^+11j^​+11k^, 92i^+βj^−8k^\frac{9}{2}\hat{i}+\beta\hat{j}-8\hat{k}29​i^+βj^​−8k^ are collinear, then (19α−6β)2(19\alpha-6\beta)^{2}(19α−6β)2 is equal to
  1. (A)36
  2. (B)16
  3. (C)25
  4. (D)49

Correct answer: (A)

Step-by-step solution →
Q111·MathematicsSingle correctJEE Main 2023
The area of the quadrilateral ABCD with vertices A(2,1,1)A(2,1,1)A(2,1,1), B(1,2,5)B(1,2,5)B(1,2,5), C(−2,−3,5)C(-2,-3,5)C(−2,−3,5) and D(1,−6,−7)D(1,-6,-7)D(1,−6,−7) is equal to
  1. (A)48
  2. (B)8388\sqrt{38}838​
  3. (C)54
  4. (D)9389\sqrt{38}938​

Correct answer: (B)

Step-by-step solution →
Q112·MathematicsSingle correctJEE Main 2023
Let the vectors a⃗,b⃗,c⃗\vec{a},\vec{b},\vec{c}a,b,c represent three coterminous edges of a parallelepiped of volume VVV. Then the volume of the parallelepiped, whose coterminous edges are represented by a⃗, b⃗+c⃗\vec{a},\ \vec{b}+\vec{c}a, b+c and a⃗+2b⃗+3c⃗\vec{a}+2\vec{b}+3\vec{c}a+2b+3c is equal to
  1. (A)3V
  2. (B)6V
  3. (C)2V
  4. (D)V

Correct answer: (B)

Step-by-step solution →
Q113·MathematicsSingle correctJEE Main 2023
The sum of all values of α\alphaα, for which the points whose position vectors i^−2j^+3k^\hat{i}-2\hat{j}+3\hat{k}i^−2j^​+3k^, 2i^−3j^+4k^2\hat{i}-3\hat{j}+4\hat{k}2i^−3j^​+4k^, (α+1)i^+2k^(\alpha+1)\hat{i}+2\hat{k}(α+1)i^+2k^ and 9i^+(α−8)j^+6k^9\hat{i}+(\alpha-8)\hat{j}+6\hat{k}9i^+(α−8)j^​+6k^ are coplanar, is equal to
  1. (A)6
  2. (B)4
  3. (C)−2-2−2
  4. (D)2

Correct answer: (D)

Step-by-step solution →
Q114·MathematicsSingle correctJEE Main 2023
Let a⃗=2i^+3j^+4k^, b⃗=2i^−2j^−2k^\vec a=2\hat i+3\hat j+4\hat k,\,\vec b=2\hat i-2\hat j-2\hat ka=2i^+3j^​+4k^,b=2i^−2j^​−2k^ and c⃗=−i^+4j^+3k^\vec c=-\hat i+4\hat j+3\hat kc=−i^+4j^​+3k^. If d⃗\vec dd is a vector perpendicular to both b⃗\vec bb and c⃗\vec cc and a⃗⋅d⃗=18\vec a\cdot\vec d=18a⋅d=18, then ∣a⃗×d⃗∣2|\vec a\times\vec d|^2∣a×d∣2 is equal to:
  1. (A)640640640
  2. (B)760760760
  3. (C)680680680
  4. (D)720720720

Correct answer: (D)

Step-by-step solution →
Q115·MathematicsSingle correctJEE Main 2023
Let the position vectors of the points A, B, CA,\,B,\,CA,B,C and DDD be 5i^+5j^+2λk^, i^+2j^+3k^, −2i^+λj^+4k^5\hat i+5\hat j+2\lambda\hat k,\,\hat i+2\hat j+3\hat k,\,-2\hat i+\lambda\hat j+4\hat k5i^+5j^​+2λk^,i^+2j^​+3k^,−2i^+λj^​+4k^ and −i^+5j^+6k^-\hat i+5\hat j+6\hat k−i^+5j^​+6k^. Let the set S={λ∈R:A,B,C and D are coplanar}S=\{\lambda\in\mathbb{R}: A,B,C\text{ and }D\text{ are coplanar}\}S={λ∈R:A,B,C and D are coplanar}. Then ∑λ∈S(λ+2)2\displaystyle\sum_{\lambda\in S}(\lambda+2)^{2}λ∈S∑​(λ+2)2 is equal to:
  1. (A)414141
  2. (B)252525
  3. (C)131313
  4. (D)372\dfrac{37}{2}237​

Correct answer: (A)

Step-by-step solution →
Q116·MathematicsNumericalJEE Main 2023
Let v⃗=αi^+2j^−3k^\vec{v}=\alpha\hat{i}+2\hat{j}-3\hat{k}v=αi^+2j^​−3k^, w⃗=2αi^+j^−k^\vec{w}=2\alpha\hat{i}+\hat{j}-\hat{k}w=2αi^+j^​−k^ and u⃗\vec{u}u be a vector such that ∣u⃗∣=α>0|\vec{u}|=\alpha>0∣u∣=α>0. If the minimum value of the scalar triple product [u⃗ v⃗ w⃗][\vec{u}\ \vec{v}\ \vec{w}][u v w] is −α3401-\alpha\sqrt{3401}−α3401​, and ∣u⃗⋅i^∣2=mn|\vec{u}\cdot\hat{i}|^2=\frac{m}{n}∣u⋅i^∣2=nm​ where mmm and nnn are coprime natural numbers, then m+nm+nm+n is equal to _______ .

Correct answer: 3501

Step-by-step solution →
Q117·MathematicsSingle correctJEE Main 2023
Let a⃗=5i^−j^−3k^\vec a=5\hat i-\hat j-3\hat ka=5i^−j^​−3k^ and b⃗=i^+3j^+5k^\vec b=\hat i+3\hat j+5\hat kb=i^+3j^​+5k^ be two vectors. Then which one of the following statements is TRUE?
  1. (A)Projection of a⃗\vec aa on b⃗\vec bb is 1335\dfrac{13}{\sqrt{35}}35​13​ and the direction of the projection vector is same as of b⃗\vec bb.
  2. (B)Projection of a⃗\vec aa on b⃗\vec bb is 1335\dfrac{13}{\sqrt{35}}35​13​ and the direction of the projection vector is opposite to the direction of b⃗\vec bb.
  3. (C)Projection of a⃗\vec aa on b⃗\vec bb is −1335\dfrac{-13}{\sqrt{35}}35​−13​ and the direction of the projection vector is same as of b⃗\vec bb.
  4. (D)Projection of a⃗\vec aa on b⃗\vec bb is −1335\dfrac{-13}{\sqrt{35}}35​−13​ and the direction of the projection vector is opposite to the direction of b⃗\vec bb.

Correct answer: (D)

Step-by-step solution →
Q118·MathematicsSingle correctJEE Main 2023
Let a⃗=2i^−7j^+5k^\vec a=2\hat i-7\hat j+5\hat ka=2i^−7j^​+5k^, b⃗=i^+k^\vec b=\hat i+\hat kb=i^+k^ and c⃗=i^+2j^−3k^\vec c=\hat i+2\hat j-3\hat kc=i^+2j^​−3k^ be three given vectors. If r⃗\vec rr is a vector such that r⃗×a⃗=c⃗×a⃗\vec r\times\vec a=\vec c\times\vec ar×a=c×a and r⃗⋅b⃗=0\vec r\cdot\vec b=0r⋅b=0, then ∣r⃗∣|\vec r|∣r∣ is equal to:
  1. (A)1152\dfrac{11}{5}\sqrt2511​2​
  2. (B)9147\dfrac{\sqrt{914}}{7}7914​​
  3. (C)1172\dfrac{11}{7}\sqrt2711​2​
  4. (D)117\dfrac{11}{7}711​

Correct answer: (C)

Step-by-step solution →
Q119·MathematicsNumericalJEE Main 2023
A(2,6,2)A(2,6,2)A(2,6,2), B(−4,0,λ)B(-4,0,\lambda)B(−4,0,λ), C(2,3,−1)C(2,3,-1)C(2,3,−1) and D(4,5,0)D(4,5,0)D(4,5,0), ∣λ∣≤5|\lambda|\le 5∣λ∣≤5 are the vertices of a quadrilateral ABCDABCDABCD. If its area is 181818 square units, then 5−6λ5-6\lambda5−6λ is equal to _______ .

Correct answer: 11

Step-by-step solution →
Q120·MathematicsSingle correctJEE Main 2023
Let a⃗=i^+2j^+3k^, b⃗=i^−j^+2k^\vec a=\hat i+2\hat j+3\hat k,\ \vec b=\hat i-\hat j+2\hat ka=i^+2j^​+3k^, b=i^−j^​+2k^ and c⃗=5i^−3j^+3k^\vec c=5\hat i-3\hat j+3\hat kc=5i^−3j^​+3k^ be three vectors. If r⃗\vec rr is a vector such that r⃗×b⃗=c⃗×b⃗\vec r\times\vec b=\vec c\times\vec br×b=c×b and r⃗⋅a⃗=0\vec r\cdot\vec a=0r⋅a=0, then 25∣r⃗∣225|\vec r|^225∣r∣2 is equal to
  1. (A)560560560
  2. (B)449449449
  3. (C)339339339
  4. (D)336336336

Correct answer: (C)

Step-by-step solution →
Q121·MathematicsNumericalJEE Main 2023
Let a⃗,b⃗,c⃗\vec a,\vec b,\vec ca,b,c be three vectors such that ∣a⃗∣=31, 4∣b⃗∣=∣c⃗∣=2|\vec a|=\sqrt{31},\ 4|\vec b|=|\vec c|=2∣a∣=31​, 4∣b∣=∣c∣=2 and 2(a⃗×b⃗)=3(c⃗×a⃗)2(\vec a\times\vec b)=3(\vec c\times\vec a)2(a×b)=3(c×a). If the angle between b⃗\vec bb and c⃗\vec cc is 2π3\dfrac{2\pi}{3}32π​, then (a⃗×c⃗a⃗⋅b⃗)2\left(\dfrac{\vec a\times\vec c}{\vec a\cdot\vec b}\right)^2(a⋅ba×c​)2 is equal to

Correct answer: 3

Step-by-step solution →
Q122·MathematicsSingle correctJEE Main 2023
Let a⃗=2i^+j^+k^\vec{a}=2\hat{i}+\hat{j}+\hat{k}a=2i^+j^​+k^, and b⃗\vec{b}b and c⃗\vec{c}c be two nonzero vectors such that ∣a⃗+b⃗+c⃗∣=∣a⃗+b⃗−c⃗∣|\vec{a}+\vec{b}+\vec{c}|=|\vec{a}+\vec{b}-\vec{c}|∣a+b+c∣=∣a+b−c∣ and b⃗⋅c⃗=0\vec{b}\cdot\vec{c}=0b⋅c=0. Consider the following two statements: (A) ∣a⃗+λc⃗∣≥∣a⃗∣|\vec{a}+\lambda\vec{c}|\ge|\vec{a}|∣a+λc∣≥∣a∣ for all λ∈R\lambda\in\mathbb{R}λ∈R. (B) a⃗\vec{a}a and c⃗\vec{c}c are always parallel. Then.
  1. (A)both (A) and (B) are correct
  2. (B)only (A) is correct
  3. (C)neither (A) nor (B) is correct
  4. (D)only (B) is correct

Correct answer: (B)

Step-by-step solution →
Q123·MathematicsNumericalJEE Main 2023
Let a⃗\vec{a}a and b⃗\vec{b}b be two vectors such that ∣a⃗∣=14|\vec{a}| = \sqrt{14}∣a∣=14​, ∣b⃗∣=6|\vec{b}| = \sqrt{6}∣b∣=6​ and ∣a⃗×b⃗∣=48|\vec{a} \times \vec{b}| = \sqrt{48}∣a×b∣=48​. Then (a⃗⋅b⃗)2(\vec{a} \cdot \vec{b})^2(a⋅b)2 is equal to

Correct answer: 36

Step-by-step solution →
Q124·MathematicsSingle correctJEE Main 2023
Let λ∈R\lambda \in \mathbb{R}λ∈R, a⃗=λi^+2j^−3k^\vec{a}=\lambda \hat{i} + 2\hat{j} - 3\hat{k}a=λi^+2j^​−3k^, b⃗=i^−λj^+2k^\vec{b}=\hat{i} - \lambda\hat{j} + 2\hat{k}b=i^−λj^​+2k^. If ((a⃗+b⃗)×(a⃗×b⃗))×(a⃗−b⃗)=8i^−40j^−24k^\left((\vec{a} + \vec{b}) \times (\vec{a} \times \vec{b})\right) \times (\vec{a} - \vec{b})=8\hat{i} - 40\hat{j} - 24\hat{k}((a+b)×(a×b))×(a−b)=8i^−40j^​−24k^, then ∣λ(a⃗+b⃗)×(a⃗−b⃗)∣2\left|\lambda(\vec{a} + \vec{b}) \times (\vec{a} - \vec{b})\right|^2​λ(a+b)×(a−b)​2 is equal to:
  1. (A)132132132
  2. (B)136136136
  3. (C)140140140
  4. (D)144144144

Correct answer: (C)

Step-by-step solution →
Q125·MathematicsSingle correctJEE Main 2023
Let a⃗\vec{a}a and b⃗\vec{b}b be two vectors. Let ∣a⃗∣=1|\vec{a}|=1∣a∣=1, ∣b⃗∣=4|\vec{b}|=4∣b∣=4 and a⃗⋅b⃗=2\vec{a} \cdot \vec{b}=2a⋅b=2. If c⃗=(2a⃗×b⃗)−3b⃗\vec{c}=(2\vec{a} \times \vec{b}) - 3\vec{b}c=(2a×b)−3b, then the value of b⃗⋅c⃗\vec{b} \cdot \vec{c}b⋅c is:
  1. (A)−24-24−24
  2. (B)−84-84−84
  3. (C)−48-48−48
  4. (D)−60-60−60

Correct answer: (C)

Step-by-step solution →
Q126·MathematicsSingle correctJEE Main 2023
If a⃗,b⃗,c⃗\vec a,\vec b,\vec ca,b,c are three non-zero vectors and n^\hat nn^ is a unit vector perpendicular to c⃗\vec cc such that a⃗=αb⃗−n^\vec a=\alpha\vec b-\hat na=αb−n^, (α≠0)(\alpha\neq0)(α=0) and b⃗⋅c⃗=12\vec b\cdot\vec c=12b⋅c=12, then ∣c⃗×(a⃗×b⃗)∣|\vec c\times(\vec a\times\vec b)|∣c×(a×b)∣ is equal to:
  1. (A)999
  2. (B)151515
  3. (C)666
  4. (D)121212

Correct answer: (D)

Step-by-step solution →
Q127·MathematicsSingle correctJEE Main 2023
If a⃗=i^+2k^, b⃗=i^+j^+k^, c⃗=7i^−3j^+4k^, r⃗×b⃗+b⃗×c⃗=0⃗\vec a=\hat i+2\hat k,\,\vec b=\hat i+\hat j+\hat k,\,\vec c=7\hat i-3\hat j+4\hat k,\,\vec r\times\vec b+\vec b\times\vec c=\vec 0a=i^+2k^,b=i^+j^​+k^,c=7i^−3j^​+4k^,r×b+b×c=0 and r⃗⋅a⃗=0\vec r\cdot\vec a=0r⋅a=0. Then r⃗⋅c⃗\vec r\cdot\vec cr⋅c is equal to:
  1. (A)323232
  2. (B)303030
  3. (C)363636
  4. (D)343434

Correct answer: (D)

Step-by-step solution →
Q128·MathematicsSingle correctJEE Main 2023
If the vectors a⃗=λi^+μj^+4k^\vec{a} = \lambda\hat{i} + \mu\hat{j} + 4\hat{k}a=λi^+μj^​+4k^, b⃗=2i^+4j^−2k^\vec{b} = 2\hat{i} + 4\hat{j} - 2\hat{k}b=2i^+4j^​−2k^ and c⃗=2i^+3j^+k^\vec{c} = 2\hat{i} + 3\hat{j} + \hat{k}c=2i^+3j^​+k^ are coplanar and the projection of a⃗\vec{a}a on the vector b⃗\vec{b}b is 54\sqrt{54}54​ units, then the sum of all possible values of λ+μ\lambda + \muλ+μ is equal to
  1. (A)0
  2. (B)24
  3. (C)6
  4. (D)18

Correct answer: (B)

Step-by-step solution →
Q129·MathematicsNumericalJEE Main 2023
Let a⃗\vec{a}a, b⃗\vec{b}b and c⃗\vec{c}c be three non-zero non-coplanar vectors. Let the position vectors of four points AAA, BBB, CCC and DDD be a⃗−b⃗+c⃗\vec{a} - \vec{b} + \vec{c}a−b+c, λa⃗−3b⃗+4c⃗\lambda\vec{a} - 3\vec{b} + 4\vec{c}λa−3b+4c, −a⃗+2b⃗−3c⃗-\vec{a} + 2\vec{b} - 3\vec{c}−a+2b−3c and 2a⃗−4b⃗+6c⃗2\vec{a} - 4\vec{b} + 6\vec{c}2a−4b+6c respectively. If AB⃗\vec{AB}AB, AC⃗\vec{AC}AC and AD⃗\vec{AD}AD are coplanar, then λ\lambdaλ is equal to ________ .

Correct answer: 2

Step-by-step solution →
Q130·MathematicsSingle correctJEE Main 2023
Let a⃗=4i^+3j^\vec a=4\hat i+3\hat ja=4i^+3j^​ and b⃗=3i^−4j^+5k^\vec b=3\hat i-4\hat j+5\hat kb=3i^−4j^​+5k^. If c⃗\vec cc is a vector such that c⃗⋅(a⃗×b⃗)+25=0, c⃗⋅(i^+j^+k^)=4\vec c\cdot(\vec a\times\vec b)+25=0,\,\vec c\cdot(\hat i+\hat j+\hat k)=4c⋅(a×b)+25=0,c⋅(i^+j^​+k^)=4, and projection of c⃗\vec cc on a⃗\vec aa is 111, then the projection of c⃗\vec cc on b⃗\vec bb equals:
  1. (A)15\dfrac1551​
  2. (B)52\dfrac{5}{\sqrt2}2​5​
  3. (C)32\dfrac{3}{\sqrt2}2​3​
  4. (D)12\dfrac{1}{\sqrt2}2​1​

Correct answer: (B)

Step-by-step solution →
Q131·MathematicsSingle correctJEE Main 2023
Let a⃗,b⃗\vec{a},\vec{b}a,b and c⃗\vec{c}c be three non zero vectors such that b⃗⋅c⃗=0\vec{b}\cdot\vec{c}=0b⋅c=0 and a⃗×(b⃗×c⃗)=b⃗−c⃗2\vec{a}\times(\vec{b}\times\vec{c})=\dfrac{\vec{b}-\vec{c}}{2}a×(b×c)=2b−c​. If d⃗\vec{d}d be a vector such that b⃗⋅d⃗=a⃗⋅b⃗\vec{b}\cdot\vec{d}=\vec{a}\cdot\vec{b}b⋅d=a⋅b, then (a⃗×b⃗)⋅(c⃗×d⃗)(\vec{a}\times\vec{b})\cdot(\vec{c}\times\vec{d})(a×b)⋅(c×d) is equal to:
  1. (A)−14-\dfrac{1}{4}−41​
  2. (B)14\dfrac{1}{4}41​
  3. (C)34\dfrac{3}{4}43​
  4. (D)12\dfrac{1}{2}21​

Correct answer: (B)

Step-by-step solution →
Q132·MathematicsSingle correctJEE Main 2023
Let a⃗=−i^−j^+k^\vec{a} = -\hat{i} - \hat{j} + \hat{k}a=−i^−j^​+k^, a⃗⋅b⃗=1\vec{a} \cdot \vec{b} = 1a⋅b=1 and a⃗×b⃗=i^−j^\vec{a} \times \vec{b} = \hat{i} - \hat{j}a×b=i^−j^​. Then a⃗−6b⃗\vec{a} - 6\vec{b}a−6b is equal to
  1. (A)3(i^−j^+k^)3(\hat{i} - \hat{j} + \hat{k})3(i^−j^​+k^)
  2. (B)(i^+j^−k^)(\hat{i} + \hat{j} - \hat{k})(i^+j^​−k^)
  3. (C)3(i^+j^+k^)3(\hat{i} + \hat{j} + \hat{k})3(i^+j^​+k^)
  4. (D)3(i^−j^−k^)3(\hat{i} - \hat{j} - \hat{k})3(i^−j^​−k^)

Correct answer: (C)

Step-by-step solution →
Q133·MathematicsSingle correctJEE Main 2023
The vector a⃗=−i^+2j^+k^\vec{a}=-\hat{i}+2\hat{j}+\hat{k}a=−i^+2j^​+k^ is rotated through a right angle, passing through the yyy-axis in its way and the resulting vector is b⃗\vec{b}b. Then the projection of 3a⃗+2 b⃗3\vec{a}+\sqrt{2}\,\vec{b}3a+2​b on c⃗=5i^+4j^+3k^\vec{c}=5\hat{i}+4\hat{j}+3\hat{k}c=5i^+4j^​+3k^ is:
  1. (A)232\sqrt{3}23​
  2. (B)111
  3. (C)323\sqrt{2}32​
  4. (D)6\sqrt{6}6​

Correct answer: (C)

Step-by-step solution →
Q134·MathematicsSingle correctJEE Main 2023
If the four points, whose position vectors are 3i^−4j^+2k^3\hat{i} - 4\hat{j} + 2\hat{k}3i^−4j^​+2k^, i^+2j^−k^\hat{i} + 2\hat{j} - \hat{k}i^+2j^​−k^, −2i^−j^+3k^-2\hat{i} - \hat{j} + 3\hat{k}−2i^−j^​+3k^ and 5i^−2αj^+4k^5\hat{i} - 2\alpha\hat{j} + 4\hat{k}5i^−2αj^​+4k^ are coplanar, then α\alphaα is equal to
  1. (A)7317\dfrac{73}{17}1773​
  2. (B)10717\dfrac{107}{17}17107​
  3. (C)−7317\dfrac{-73}{17}17−73​
  4. (D)−10717\dfrac{-107}{17}17−107​

Correct answer: (A)

Step-by-step solution →
Q135·MathematicsSingle correctJEE Main 2023
Let PQRPQRPQR be a triangle. The points AAA, BBB and CCC are on the sides QRQRQR, RPRPRP and PQPQPQ respectively such that QAAR=RBBP=PCCQ=12\dfrac{QA}{AR}=\dfrac{RB}{BP}=\dfrac{PC}{CQ}=\dfrac{1}{2}ARQA​=BPRB​=CQPC​=21​. Then Area(△PQR)Area(△ABC)\dfrac{\text{Area}(\triangle PQR)}{\text{Area}(\triangle ABC)}Area(△ABC)Area(△PQR)​ is equal to
  1. (A)444
  2. (B)333
  3. (C)52\tfrac{5}{2}25​
  4. (D)222

Correct answer: (B)

Step-by-step solution →
Q136·MathematicsSingle correctJEE Main 2023
Let α⃗=4i^+3j^+5k^\vec\alpha=4\hat i+3\hat j+5\hat kα=4i^+3j^​+5k^ and β⃗=i^+2j^−4k^\vec\beta=\hat i+2\hat j-4\hat kβ​=i^+2j^​−4k^. Let β1⃗\vec{\beta_1}β1​​ be parallel to α⃗\vec\alphaα and β2⃗\vec{\beta_2}β2​​ be perpendicular to α⃗\vec\alphaα. If β⃗=β1⃗+β2⃗\vec\beta=\vec{\beta_1}+\vec{\beta_2}β​=β1​​+β2​​, then the value of 5β2⃗⋅(i^+j^+k^)5\vec{\beta_2}\cdot(\hat i+\hat j+\hat k)5β2​​⋅(i^+j^​+k^) is
  1. (A)777
  2. (B)999
  3. (C)666
  4. (D)111111

Correct answer: (A)

Step-by-step solution →
Q137·MathematicsSingle correctJEE Main 2023
Let u⃗=i^−j^−2k^\vec{u}=\hat{i}-\hat{j}-2\hat{k}u=i^−j^​−2k^, v⃗=2i^+j^−k^\vec{v}=2\hat{i}+\hat{j}-\hat{k}v=2i^+j^​−k^, v⃗⋅w⃗=2\vec{v}\cdot\vec{w}=2v⋅w=2 and v⃗×w⃗=u⃗+λv⃗\vec{v}\times\vec{w}=\vec{u}+\lambda\vec{v}v×w=u+λv. Then u⃗⋅w⃗\vec{u}\cdot\vec{w}u⋅w is equal to
  1. (A)222
  2. (B)32\tfrac{3}{2}23​
  3. (C)111
  4. (D)−23-\tfrac{2}{3}−32​

Correct answer: (C)

Step-by-step solution →
Q138·MathematicsNumericalJEE Main 2023
Let a⃗=i^+2j^+λk^, b⃗=3i^−5j^−λk^, a⃗⋅c⃗=7, 2b⃗⋅c⃗+43=0, a⃗×c⃗=b⃗×c⃗\vec a=\hat i+2\hat j+\lambda\hat k,\ \vec b=3\hat i-5\hat j-\lambda\hat k,\ \vec a\cdot\vec c=7,\ 2\vec b\cdot\vec c+43=0,\ \vec a\times\vec c=\vec b\times\vec ca=i^+2j^​+λk^, b=3i^−5j^​−λk^, a⋅c=7, 2b⋅c+43=0, a×c=b×c. Then ∣a⃗⋅b⃗∣|\vec a\cdot\vec b|∣a⋅b∣ is equal to

Correct answer: 8

Step-by-step solution →
Q139·MathematicsMultiple correctJEE Advanced 2022
Let i^\hat{i}i^, j^\hat{j}j^​ and k^\hat{k}k^ be the unit vectors along the three positive coordinate axes. Let a⃗=3i^+j^−k^\vec{a} = 3\hat{i} + \hat{j} - \hat{k}a=3i^+j^​−k^, b⃗=i^+b2j^+b3k^\vec{b} = \hat{i} + b_{2}\hat{j} + b_{3}\hat{k}b=i^+b2​j^​+b3​k^, b2_{2}2​, b3_{3}3​ ∈ R c⃗=c1i^+c2j^+c3k^\vec{c} = c_{1}\hat{i} + c_{2}\hat{j} + c_{3}\hat{k}c=c1​i^+c2​j^​+c3​k^, c1_{1}1​, c2_{2}2​, c3_{3}3​ ∈ R be the vectors such that b2_{2}2​ b3_{3}3​ > 0, a⃗⋅b⃗=0\vec{a} \cdot \vec{b} = 0a⋅b=0 and (0−c3c2c30−c1−c2c10)(1b2b3)=(3−c11−c2−1−c3)\begin{pmatrix} 0 & -c_{3} & c_{2} \\ c_{3} & 0 & -c_{1} \\ -c_{2} & c_{1} & 0 \end{pmatrix}\begin{pmatrix} 1 \\ b_{2} \\ b_{3} \end{pmatrix} = \begin{pmatrix} 3-c_{1} \\ 1-c_{2} \\ -1-c_{3} \end{pmatrix}​0c3​−c2​​−c3​0c1​​c2​−c1​0​​​1b2​b3​​​=​3−c1​1−c2​−1−c3​​​ Then, which of the following is/are TRUE?
  1. (A)a⃗⋅c⃗=0\vec{a} \cdot \vec{c} = 0a⋅c=0
  2. (B)b⃗⋅c⃗=0\vec{b} \cdot \vec{c} = 0b⋅c=0
  3. (C)∣b⃗∣>10\left|\vec{b}\right| > \sqrt{10}​b​>10​
  4. (D)∣c⃗∣≤11\left|\vec{c}\right| \leq \sqrt{11}∣c∣≤11​ .

Correct answer: (B), (C), (D)

Step-by-step solution →
Q140·MathematicsNumericalJEE Main 2022
Let a⃗\vec{a}a and b⃗\vec{b}b be two vectors such that ∣a⃗+b⃗∣2=∣a⃗∣2+2∣b⃗∣2|\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + 2|\vec{b}|^2∣a+b∣2=∣a∣2+2∣b∣2, a⃗⋅b⃗=3\vec{a} \cdot \vec{b} = 3a⋅b=3 and ∣a⃗×b⃗∣2=75|\vec{a} \times \vec{b}|^2 = 75∣a×b∣2=75. Then ∣a⃗∣2|\vec{a}|^2∣a∣2 is equal to____.

Correct answer: 14

Step-by-step solution →
Q141·MathematicsSingle correctJEE Main 2022
Let a⃗=3i^+j^\vec{a} = 3\hat{i} + \hat{j}a=3i^+j^​ and b⃗=i^+2j^+k^\vec{b} = \hat{i} + 2\hat{j} + \hat{k}b=i^+2j^​+k^. Let c⃗\vec{c}c be a vector satisfying a⃗×(b⃗×c⃗)=b⃗+λc⃗\vec{a} \times (\vec{b} \times \vec{c}) = \vec{b} + \lambda\vec{c}a×(b×c)=b+λc. If b⃗\vec{b}b and c⃗\vec{c}c are non-parallel, then the value of λ\lambdaλ is :
  1. (A)−5-5−5
  2. (B)5
  3. (C)1
  4. (D)−1-1−1

Correct answer: (A)

Step-by-step solution →
Q142·MathematicsSingle correctJEE Main 2022
Let a^\hat{a}a^ and b^\hat{b}b^ be two unit vectors such that the angle between them is π4\frac{\pi}{4}4π​. If θ\thetaθ is the angle between the vectors (a^+b^)(\hat{a} + \hat{b})(a^+b^) and (a^+2b^+2(a^×b^))(\hat{a} + 2\hat{b} + 2(\hat{a} \times \hat{b}))(a^+2b^+2(a^×b^)), then the value of 164cos⁡2θ164 \cos^{2}\theta164cos2θ is equal to :
  1. (A)90+27290 + 27\sqrt{2}90+272​
  2. (B)45+18245 + 18\sqrt{2}45+182​
  3. (C)90+3290 + 3\sqrt{2}90+32​
  4. (D)54+90254 + 90\sqrt{2}54+902​

Correct answer: (A)

Step-by-step solution →
Q143·MathematicsSingle correctJEE Main 2022
Let a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c}a,b,c be three coplanar concurrent vectors such that angles between any two of them is same. If the product of their magnitudes is 14 and (a⃗×b⃗)⋅(b⃗×c⃗)+(b⃗×c⃗)⋅(c⃗×a⃗)+(c⃗×a⃗)⋅(a⃗×b⃗)=168\left(\vec{a} \times \vec{b}\right) \cdot \left(\vec{b} \times \vec{c}\right) + \left(\vec{b} \times \vec{c}\right) \cdot \left(\vec{c} \times \vec{a}\right) + \left(\vec{c} \times \vec{a}\right) \cdot \left(\vec{a} \times \vec{b}\right) = 168(a×b)⋅(b×c)+(b×c)⋅(c×a)+(c×a)⋅(a×b)=168 then ∣a⃗∣+∣b⃗∣+∣c⃗∣|\vec{a}| + |\vec{b}| + |\vec{c}|∣a∣+∣b∣+∣c∣ is equal to:
  1. (A)10
  2. (B)14
  3. (C)16
  4. (D)18

Correct answer: (C)

Step-by-step solution →
Q144·MathematicsSingle correctJEE Main 2022
Let the vectors a⃗=(1+t)i^+(1−t)j^+k^\vec{a} = (1+t)\hat{i} + (1-t)\hat{j} + \hat{k}a=(1+t)i^+(1−t)j^​+k^, b⃗=(1−t)i^+(1+t)j^+2k^\vec{b} = (1-t)\hat{i} + (1+t)\hat{j} + 2\hat{k}b=(1−t)i^+(1+t)j^​+2k^ and c⃗=ti^−tj^+k^\vec{c} = t\hat{i} - t\hat{j} + \hat{k}c=ti^−tj^​+k^, t∈Rt \in Rt∈R be such that for α,β,γ∈R\alpha, \beta, \gamma \in Rα,β,γ∈R, αa⃗+βb⃗+γc⃗=0⃗⇒α=β=γ=0\alpha\vec{a} + \beta\vec{b} + \gamma\vec{c} = \vec{0} \Rightarrow \alpha = \beta = \gamma = 0αa+βb+γc=0⇒α=β=γ=0. Then, the set of all values of t is :
  1. (A)a non-empty finite set
  2. (B)equal to N
  3. (C)equal to R − {0}
  4. (D)equal to R

Correct answer: (C)

Step-by-step solution →
Q145·MathematicsSingle correctJEE Main 2022
Let S be the set of all a∈Ra \in Ra∈R for which the angle between the vectors u⃗=a(log⁡eb)i^−6j^+3k^\vec{u}=a\left(\log_{e}b\right)\hat{i}-6\hat{j}+3\hat{k}u=a(loge​b)i^−6j^​+3k^ and v⃗=(log⁡eb)i^+2j^+2a(log⁡eb)k^,(b>1)\vec{v}=\left(\log_{e}b\right)\hat{i}+2\hat{j}+2a\left(\log_{e}b\right)\hat{k}, (b>1)v=(loge​b)i^+2j^​+2a(loge​b)k^,(b>1) is acute. Then S is equal to
  1. (A)(−∞,−43)\left(-\infty,-\frac{4}{3}\right)(−∞,−34​)
  2. (B)Φ\PhiΦ
  3. (C)(−43,0)\left(-\frac{4}{3},0\right)(−34​,0)
  4. (D)(127,∞)\left(\frac{12}{7},\infty\right)(712​,∞)

Correct answer: (C)

Step-by-step solution →
Q146·MathematicsSingle correctJEE Main 2022
Let a vector a⃗\vec{a}a has a magnitude 9. Let a vector b⃗\vec{b}b be such that for every (x,y)∈R×R−{(0,0)}(x, y) \in R \times R - \{(0,0)\}(x,y)∈R×R−{(0,0)}, the vector (xa⃗+yb⃗)(x\vec{a} + y\vec{b})(xa+yb) is perpendicular to the vector (6y a⃗−18x b⃗)(6y\,\vec{a} - 18x\,\vec{b})(6ya−18xb). Then the value of ∣a⃗×b⃗∣\left|\vec{a} \times \vec{b}\right|​a×b​ is equal to:
  1. (A)939\sqrt{3}93​
  2. (B)27327\sqrt{3}273​
  3. (C)9
  4. (D)81

Correct answer: (B)

Step-by-step solution →
Q147·MathematicsSingle correctJEE Main 2022
Let a⃗=αi^+j^+βk^\vec{a} = \alpha\hat{i} + \hat{j} + \beta\hat{k}a=αi^+j^​+βk^ and b⃗=3i^−5j^+4k^\vec{b} = 3\hat{i} - 5\hat{j} + 4\hat{k}b=3i^−5j^​+4k^ be two vectors, such that a⃗×b⃗=−i^+9i^+12k\vec{a} \times \vec{b} = -\hat{i} + 9\hat{i} + 12ka×b=−i^+9i^+12k. Then the projection of b⃗−2a⃗\vec{b} - 2\vec{a}b−2a on b⃗+a⃗\vec{b} + \vec{a}b+a is equal to
  1. (A)2
  2. (B)395\frac{39}{5}539​
  3. (C)9
  4. (D)465\frac{46}{5}546​

Correct answer: (D)

Step-by-step solution →
Q148·MathematicsSingle correctJEE Main 2022
Let a⃗=2i^−j^+5k^\vec{a} = 2\hat{i} - \hat{j} + 5\hat{k}a=2i^−j^​+5k^ and b⃗=αi^+βj^+2k^\vec{b} = \alpha\hat{i} + \beta\hat{j} + 2\hat{k}b=αi^+βj^​+2k^. If ((a⃗×b⃗)×i^).k^=232\left(\left(\vec{a} \times \vec{b}\right) \times \hat{i}\right).\hat{k} = \frac{23}{2}((a×b)×i^).k^=223​, then ∣b⃗×2j^∣\left| \vec{b} \times 2\hat{j} \right|​b×2j^​​ is equal to
  1. (A)4
  2. (B)5
  3. (C)21\sqrt{21}21​
  4. (D)17\sqrt{17}17​

Correct answer: (B)

Step-by-step solution →
Q149·MathematicsSingle correctJEE Main 2022
Let a⃗=αi^+j^−k^\vec{a} = \alpha\hat{i} + \hat{j} - \hat{k}a=αi^+j^​−k^ and b⃗=2i^+j^−αk^,α>0\vec{b} = 2\hat{i} + \hat{j} - \alpha\hat{k}, \alpha > 0b=2i^+j^​−αk^,α>0. If the projection of a⃗×b⃗\vec{a} \times \vec{b}a×b on the vector −i^+2j^−2k^-\hat{i} + 2\hat{j} - 2\hat{k}−i^+2j^​−2k^ is 30, then α is equal to
  1. (A)152\dfrac{15}{2}215​
  2. (B)8
  3. (C)132\dfrac{13}{2}213​
  4. (D)7

Correct answer: (D)

Step-by-step solution →
Q150·MathematicsSingle correctJEE Main 2022
A vector a⃗\vec{a}a is parallel to the line of intersection of the plane determined by the vectors i^,i^+j^\hat{i},\hat{i}+\hat{j}i^,i^+j^​ and the plane determined by the vectors i^−j^,i^+k^\hat{i}-\hat{j},\hat{i}+\hat{k}i^−j^​,i^+k^. The obtuse angle between a⃗\vec{a}a and the vector b⃗=i^−2j^+2k^\vec{b}=\hat{i}-2\hat{j}+2\hat{k}b=i^−2j^​+2k^ is
  1. (A)3π4\frac{3\pi}{4}43π​
  2. (B)2π3\frac{2\pi}{3}32π​
  3. (C)4π5\frac{4\pi}{5}54π​
  4. (D)5π6\frac{5\pi}{6}65π​

Correct answer: (A)

Step-by-step solution →
Q151·MathematicsSingle correctJEE Main 2022
Let ABC be a triangle such that BC→=a⃗\overrightarrow{BC} = \vec{a}BC=a, CA→=b⃗\overrightarrow{CA} = \vec{b}CA=b, AB→=c⃗\overrightarrow{AB} = \vec{c}AB=c, ∣a⃗∣=62|\vec{a}| = 6\sqrt{2}∣a∣=62​, ∣b⃗∣=23|\vec{b}| = 2\sqrt{3}∣b∣=23​ and b⃗⋅c⃗=12\vec{b} \cdot \vec{c} = 12b⋅c=12. Consider the statements : (S1) : ∣(a⃗×b⃗)+(c⃗×b⃗)∣−∣c⃗∣=6(22−1)|(\vec{a} \times \vec{b}) + (\vec{c} \times \vec{b})| - |\vec{c}| = 6(2\sqrt{2} - 1)∣(a×b)+(c×b)∣−∣c∣=6(22​−1) (S2) : ∠ABC=cos⁡−1(23)\angle ABC = \cos^{-1}\left( \sqrt{\frac{2}{3}} \right)∠ABC=cos−1(32​​). Then
  1. (A)both (S1) and (S2) are true
  2. (B)only (S1) is true
  3. (C)only (S2) is true
  4. (D)both (S1) and (S2) are false

Correct answer: (D)

Step-by-step solution →
Q152·MathematicsSingle correctJEE Main 2022
Let a⃗=αi^+3j^−k^,b⃗=3i^−βj^+4k^\vec{a} = \alpha\hat{i} + 3\hat{j} - \hat{k}, \vec{b} = 3\hat{i} - \beta\hat{j} + 4\hat{k}a=αi^+3j^​−k^,b=3i^−βj^​+4k^ and c⃗=i^+2j^−2k^\vec{c} = \hat{i} + 2\hat{j} - 2\hat{k}c=i^+2j^​−2k^ where α, β ∈ R, be three vectors. If the projection of a⃗\vec{a}a on c⃗\vec{c}c is 103\frac{10}{3}310​ and b⃗×c⃗=−6i^+10j^+7k^\vec{b} \times \vec{c} = -6\hat{i} + 10\hat{j} + 7\hat{k}b×c=−6i^+10j^​+7k^, then the value of α + β equal to :
  1. (A)3
  2. (B)4
  3. (C)5
  4. (D)6

Correct answer: (A)

Step-by-step solution →
Q153·MathematicsSingle correctJEE Main 2022
Let A, B, C be three points whose position vectors respectively are: a⃗=i^+4j^+3k^\vec{a} = \hat{i} + 4\hat{j} + 3\hat{k}a=i^+4j^​+3k^ b⃗=2i^+αj^+4k^,α∈R\vec{b} = 2\hat{i} + \alpha\hat{j} + 4\hat{k}, \alpha \in \mathbb{R}b=2i^+αj^​+4k^,α∈R c⃗=3i^−2j^+5k^\vec{c} = 3\hat{i} - 2\hat{j} + 5\hat{k}c=3i^−2j^​+5k^ If α\alphaα is the smallest positive integer for which a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c}a,b,c are non-collinear, then the length of the median, in △ABC\triangle ABC△ABC, through A is:
  1. (A)822\frac{\sqrt{82}}{2}282​​
  2. (B)622\frac{\sqrt{62}}{2}262​​
  3. (C)692\frac{\sqrt{69}}{2}269​​
  4. (D)662\frac{\sqrt{66}}{2}266​​

Correct answer: (A)

Step-by-step solution →
Q154·MathematicsSingle correctJEE Main 2022
Let a⃗=αi^+2j^−k^\vec{a} = \alpha\hat{i} + 2\hat{j} - \hat{k}a=αi^+2j^​−k^ and b⃗=−2i^+αj^+k^\vec{b} = -2\hat{i} + \alpha\hat{j} + \hat{k}b=−2i^+αj^​+k^, where α∈R\alpha \in \mathbb{R}α∈R. If the area of the parallelogram whose adjacent sides are represented by the vectors a⃗\vec{a}a and b⃗\vec{b}b is 15(α2+4)\sqrt{15\left(\alpha^{2} + 4\right)}15(α2+4)​, then the value of 2∣a⃗∣2+(a⃗⋅b⃗)∣b⃗∣22\left|\vec{a}\right|^{2} + \left(\vec{a} \cdot \vec{b}\right)\left|\vec{b}\right|^{2}2∣a∣2+(a⋅b)​b​2 is equal to
  1. (A)10
  2. (B)7
  3. (C)9
  4. (D)14

Correct answer: (D)

Step-by-step solution →
Q155·MathematicsSingle correctJEE Main 2022
Let a⃗\vec{a}a be a vector which is perpendicular to the vector 3i^+12j^+2k^3\hat{i} + \frac{1}{2}\hat{j} + 2\hat{k}3i^+21​j^​+2k^. If a⃗×(2i^+k^)=2i^−13j^−4k^\vec{a} \times \left(2\hat{i} + \hat{k}\right) = 2\hat{i} - 13\hat{j} - 4\hat{k}a×(2i^+k^)=2i^−13j^​−4k^, then the projection of the vector a⃗\vec{a}a on the vector 2i^+2j^+k^2\hat{i} + 2\hat{j} + \hat{k}2i^+2j^​+k^ is
  1. (A)13\frac{1}{3}31​
  2. (B)111
  3. (C)53\frac{5}{3}35​
  4. (D)73\frac{7}{3}37​

Correct answer: (C)

Step-by-step solution →
Q156·MathematicsNumericalJEE Main 2022
If a⃗=2i^+j^+3k^,\vec{a} = 2\hat{i} + \hat{j} + 3\hat{k},a=2i^+j^​+3k^, b⃗=3i^+3j^+k^\vec{b} = 3\hat{i} + 3\hat{j} + \hat{k}b=3i^+3j^​+k^ and c⃗=c1i^+c2j^+c3k^\vec{c} = c_1\hat{i} + c_2\hat{j} + c_3\hat{k}c=c1​i^+c2​j^​+c3​k^ are coplanar vectors and a⃗⋅c⃗=5,\vec{a} \cdot \vec{c} = 5,a⋅c=5, b⃗⊥c⃗,\vec{b} \perp \vec{c},b⊥c, then 122 (c1+c2+c3)122\,(c_1 + c_2 + c_3)122(c1​+c2​+c3​) is equal to ______.

Correct answer: 150

Step-by-step solution →
Q157·MathematicsSingle correctJEE Main 2022
Let a⃗\vec{a}a and b⃗\vec{b}b be the vectors along the diagonal of a parallelogram having area 222\sqrt{2}22​. Let the angle between a⃗\vec{a}a and b⃗\vec{b}b be acute. ∣a⃗∣=1|\vec{a}|=1∣a∣=1 and ∣a⃗⋅b⃗∣=∣a⃗×b⃗∣|\vec{a}\cdot\vec{b}|=|\vec{a}\times\vec{b}|∣a⋅b∣=∣a×b∣. If c⃗=22(a⃗×b⃗)−2b⃗\vec{c}=2\sqrt{2}\left(\vec{a}\times\vec{b}\right)-2\vec{b}c=22​(a×b)−2b, then an angle between b⃗\vec{b}b and c⃗\vec{c}c is :
  1. (A)π4\dfrac{\pi}{4}4π​
  2. (B)−π4-\dfrac{\pi}{4}−4π​
  3. (C)5π6\dfrac{5\pi}{6}65π​
  4. (D)3π4\dfrac{3\pi}{4}43π​

Correct answer: (D)

Step-by-step solution →
Q158·MathematicsSingle correctJEE Main 2022
Let a⃗=i^+j^−k^\vec{a} = \hat{i} + \hat{j} - \hat{k}a=i^+j^​−k^ and c⃗=2i^−3j^+2k^\vec{c} = 2\hat{i} - 3\hat{j} + 2\hat{k}c=2i^−3j^​+2k^. Then the number of vectors b⃗\vec{b}b such that b⃗×c⃗=a⃗\vec{b} \times \vec{c} = \vec{a}b×c=a and ∣b⃗∣∈{1,2,.....,10}\left|\vec{b}\right| \in \{1, 2, ....., 10\}​b​∈{1,2,.....,10} is :
  1. (A)0
  2. (B)1
  3. (C)2
  4. (D)3

Correct answer: (A)

Step-by-step solution →
Q159·MathematicsSingle correctJEE Main 2022
If a⃗⋅b⃗=1\vec{a}\cdot\vec{b}=1a⋅b=1, b⃗⋅c⃗=2\vec{b}\cdot\vec{c}=2b⋅c=2 and c⃗⋅a⃗=3\vec{c}\cdot\vec{a}=3c⋅a=3, then the value of [a⃗×(b⃗×c⃗), b⃗×(c⃗×a⃗), c⃗×(b⃗×a⃗)]\left[\vec{a}\times\left(\vec{b}\times\vec{c}\right),\ \vec{b}\times\left(\vec{c}\times\vec{a}\right),\ \vec{c}\times\left(\vec{b}\times\vec{a}\right)\right][a×(b×c), b×(c×a), c×(b×a)] is :
  1. (A)000
  2. (B)−6a⃗⋅(b⃗×c⃗)-6\vec{a}\cdot\left(\vec{b}\times\vec{c}\right)−6a⋅(b×c)
  3. (C)12c⃗⋅(a⃗×b⃗)12\vec{c}\cdot\left(\vec{a}\times\vec{b}\right)12c⋅(a×b)
  4. (D)−12b⃗⋅(c⃗×a⃗)-12\vec{b}\cdot\left(\vec{c}\times\vec{a}\right)−12b⋅(c×a)

Correct answer: (A)

Step-by-step solution →
Q160·MathematicsSingle correctJEE Main 2022
Let a⃗=i^+j^+2k^\vec{a}=\hat{i}+\hat{j}+2\hat{k}a=i^+j^​+2k^, b⃗=2i^−3j^+k^\vec{b}=2\hat{i}-3\hat{j}+\hat{k}b=2i^−3j^​+k^ and c⃗=i^−j^+k^\vec{c}=\hat{i}-\hat{j}+\hat{k}c=i^−j^​+k^ be three given vectors. Let v⃗\vec{v}v be a vector in the plane of a⃗\vec{a}a and b⃗\vec{b}b whose projection on c⃗\vec{c}c is 23\frac{2}{\sqrt{3}}3​2​. If v⃗⋅j^=7\vec{v}\cdot\hat{j}=7v⋅j^​=7, then v⃗⋅(i^+k^)\vec{v}\cdot\left(\hat{i}+\hat{k}\right)v⋅(i^+k^) is equal to :
  1. (A)6
  2. (B)7
  3. (C)8
  4. (D)9

Correct answer: (D)

Step-by-step solution →
Q161·MathematicsNumericalJEE Main 2022
Let θ be the angle between the vectors a⃗\vec{a}a and b⃗\vec{b}b, where ∣a⃗∣=4|\vec{a}| = 4∣a∣=4, ∣b⃗∣=3|\vec{b}| = 3∣b∣=3 θ∈(π4,π3)\theta \in \left(\frac{\pi}{4}, \frac{\pi}{3}\right)θ∈(4π​,3π​). Then ∣(a⃗−b⃗)×(a⃗+b⃗)∣2+4(a⃗⋅b⃗)2\left|\left(\vec{a} - \vec{b}\right) \times \left(\vec{a} + \vec{b}\right)\right|^2 + 4\left(\vec{a}\cdot\vec{b}\right)^2​(a−b)×(a+b)​2+4(a⋅b)2 is equal to ____

Correct answer: 576

Step-by-step solution →
Q162·MathematicsNumericalJEE Main 2022
Let b⃗=i^+j^+λk^,λ∈R\vec{b} = \hat{i} + \hat{j} + \lambda\hat{k}, \lambda \in Rb=i^+j^​+λk^,λ∈R. If a⃗\vec{a}a is a vector such that a⃗×b⃗=13i^−j^−4k^\vec{a} \times \vec{b} = 13\hat{i} - \hat{j} - 4\hat{k}a×b=13i^−j^​−4k^ and a⃗⋅b⃗+21=0\vec{a} \cdot \vec{b} + 21 = 0a⋅b+21=0, then (b⃗−a⃗)⋅(k^−j^)+(b⃗+a⃗)⋅(i^−k^)\left(\vec{b} - \vec{a}\right) \cdot \left(\hat{k} - \hat{j}\right) + \left(\vec{b} + \vec{a}\right) \cdot \left(\hat{i} - \hat{k}\right)(b−a)⋅(k^−j^​)+(b+a)⋅(i^−k^) is equal to

Correct answer: 14

Step-by-step solution →
Q163·MathematicsSingle correctJEE Main 2022
Let a⃗=a1i^+a2j^+a3k^\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}a=a1​i^+a2​j^​+a3​k^ ai>0a_i > 0ai​>0, i = 1, 2, 3 be a vector which makes equal angles with the coordinates axes OX, OY and OZ. Also, let the projection of a⃗\vec{a}a on the vector 3i^+4j^3\hat{i} + 4\hat{j}3i^+4j^​ be 7. Let b⃗\vec{b}b be a vector obtained by rotating a⃗\vec{a}a with 90°. If a⃗\vec{a}a, b⃗\vec{b}b and x-axis are coplanar, then projection of a vector b⃗\vec{b}b on 3i^+4j^3\hat{i} + 4\hat{j}3i^+4j^​ is equal to
  1. (A)7\sqrt{7}7​
  2. (B)2\sqrt{2}2​
  3. (C)222
  4. (D)777

Correct answer: (B)

Step-by-step solution →
Q164·MathematicsSingle correctJEE Main 2022
Let a^,b^\hat{a},\hat{b}a^,b^ be unit vectors. If c⃗\vec{c}c be a vector such that the angle between a^\hat{a}a^ and c⃗\vec{c}c is π12\dfrac{\pi}{12}12π​, and b^=c⃗+2(c⃗×a^)\hat{b}=\vec{c}+2(\vec{c}\times\hat{a})b^=c+2(c×a^), then ∣6c⃗∣2\left|6\vec{c}\right|^{2}∣6c∣2 is equal to
  1. (A)6(3−3)6(3-\sqrt{3})6(3−3​)
  2. (B)3+33+\sqrt{3}3+3​
  3. (C)6(3+3)6(3+\sqrt{3})6(3+3​)
  4. (D)6(3+1)6(\sqrt{3}+1)6(3​+1)

Correct answer: (C)

Step-by-step solution →
Q165·MathematicsSingle correctJEE Main 2022
Let a^\hat{a}a^ and b^\hat{b}b^ be two unit vectors such that ∣(a^+b^)+2(a^×b^)∣=2\left|\left(\hat{a}+\hat{b}\right)+2\left(\hat{a}\times\hat{b}\right)\right|=2​(a^+b^)+2(a^×b^)​=2. If θ∈(0,π)\theta \in (0, \pi)θ∈(0,π) is the angle between a^\hat{a}a^ and b^\hat{b}b^, then among the statements : (S1) : 2∣a^×b^∣=∣a^−b^∣2\left|\hat{a}\times\hat{b}\right|=\left|\hat{a}-\hat{b}\right|2​a^×b^​=​a^−b^​ (S2) : The projection of a^\hat{a}a^ on (a^+b^)\left(\hat{a}+\hat{b}\right)(a^+b^) is 12\dfrac{1}{2}21​
  1. (A)Only (S1) is true
  2. (B)Only (S2) is true
  3. (C)Both (S1) and (S2) are true
  4. (D)Both (S1) and (S2) are false

Correct answer: (C)

Step-by-step solution →
Q166·MathematicsIntegerJEE Advanced 2021
Let u⃗,v⃗\vec{u}, \vec{v}u,v and w⃗\vec{w}w be vectors in three-dimensional space, where u⃗\vec{u}u and v⃗\vec{v}v are unit vectors which are not perpendicular to each other and u⃗⋅w⃗=1\vec{u}\cdot\vec{w} = 1u⋅w=1, v⃗⋅w⃗=1\vec{v}\cdot\vec{w} = 1v⋅w=1, w⃗⋅w⃗=4\vec{w}\cdot\vec{w} = 4w⋅w=4 If the volume of the parallelopiped, whose adjacent sides are represented by the vectors u⃗,v⃗\vec{u}, \vec{v}u,v and w⃗\vec{w}w , is 2\sqrt{2}2​ , then the value of ∣3u⃗+5v⃗∣\left|3\vec{u} + 5\vec{v}\right|∣3u+5v∣ is____.

Correct answer: 7

Step-by-step solution →
Q167·MathematicsMultiple correctJEE Advanced 2021
Let O be the origin and OA⃗=2i^+2j^+k^\vec{OA} = 2\hat{i} + 2\hat{j} + \hat{k}OA=2i^+2j^​+k^, OB⃗=i^−2j^+2k^\vec{OB} = \hat{i} - 2\hat{j} + 2\hat{k}OB=i^−2j^​+2k^ and OC⃗=12(OB⃗−λOA⃗)\vec{OC} = \frac{1}{2}\left(\vec{OB} - \lambda\vec{OA}\right)OC=21​(OB−λOA) for some λ > 0. If ∣OB⃗×OC⃗∣=92\left|\vec{OB} \times \vec{OC}\right| = \frac{9}{2}​OB×OC​=29​, then which of the following statements is (are) TRUE?
  1. (A)Projection of OC⃗\vec{OC}OC on OA⃗\vec{OA}OA is −32-\frac{3}{2}−23​
  2. (B)Area of the triangle OAB is 92\frac{9}{2}29​
  3. (C)Area of the triangle ABC is 92\frac{9}{2}29​
  4. (D)The acute angle between the diagonals of the parallelogram with adjacent sides OA⃗\vec{OA}OA and OC⃗\vec{OC}OC is π3\frac{\pi}{3}3π​

Correct answer: (A), (B), (C)

Step-by-step solution →
Q168·MathematicsNumericalJEE Main 2021
Let a⃗=2i^−j^+2k^\vec{a} = 2\hat{i} - \hat{j} + 2\hat{k}a=2i^−j^​+2k^ and b⃗=i^+2j^−k^\vec{b} = \hat{i} + 2\hat{j} - \hat{k}b=i^+2j^​−k^. Let a vector v⃗\vec{v}v be in the plane containing a⃗\vec{a}a and b⃗\vec{b}b. If v⃗\vec{v}v is perpendicular to the vector 3i^+2j^−k^3\hat{i} + 2\hat{j} - \hat{k}3i^+2j^​−k^ and its projection on a⃗\vec{a}a is 19 units, then ∣2v⃗∣2\left|2\vec{v}\right|^{2}∣2v∣2 is equal to ______ .

Correct answer: 1494

Step-by-step solution →
Q169·MathematicsSingle correctJEE Main 2021
Let a⃗\vec{a}a and b⃗\vec{b}b be two vectors such that ∣2a⃗+3b⃗∣=∣3a⃗+b⃗∣\left|2\vec{a} + 3\vec{b}\right| = \left|3\vec{a} + \vec{b}\right|​2a+3b​=​3a+b​ and the angle between a⃗\vec{a}a and b⃗\vec{b}b is 60∘60^{\circ}60∘. If 18a⃗\frac{1}{8}\vec{a}81​a is a unit vector, then ∣b⃗∣\left|\vec{b}\right|​b​ is equal to :
  1. (A)4
  2. (B)6
  3. (C)5
  4. (D)8

Correct answer: (C)

Step-by-step solution →
Q170·MathematicsSingle correctJEE Main 2021
Let a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c}a,b,c be three vectors mutually perpendicular to each other and have same magnitude. If a vector r⃗\vec{r}r satisfies. a⃗×{(r⃗−b⃗)×a⃗}+b⃗×{(r⃗−c⃗)×b⃗}+c⃗×{(r⃗−a⃗)×c⃗}=0⃗\vec{a} \times \{(\vec{r} - \vec{b}) \times \vec{a}\} + \vec{b} \times \{(\vec{r} - \vec{c}) \times \vec{b}\} + \vec{c} \times \{(\vec{r} - \vec{a}) \times \vec{c}\} = \vec{0}a×{(r−b)×a}+b×{(r−c)×b}+c×{(r−a)×c}=0, then r⃗\vec{r}r is equal to :
  1. (A)13(a⃗+b⃗+c⃗)\frac{1}{3}(\vec{a} + \vec{b} + \vec{c})31​(a+b+c)
  2. (B)13(2a⃗+b⃗−c⃗)\frac{1}{3}(2\vec{a} + \vec{b} - \vec{c})31​(2a+b−c)
  3. (C)12(a⃗+b⃗+c⃗)\frac{1}{2}(\vec{a} + \vec{b} + \vec{c})21​(a+b+c)
  4. (D)12(a⃗+b⃗+2c⃗)\frac{1}{2}(\vec{a} + \vec{b} + 2\vec{c})21​(a+b+2c)

Correct answer: (C)

Step-by-step solution →
Q171·MathematicsNumericalJEE Main 2021
Let a⃗=i^+5j^+αk^\vec{a} = \hat{i} + 5\hat{j} + \alpha\hat{k}a=i^+5j^​+αk^, b⃗=i^+3j^+βk^\vec{b} = \hat{i} + 3\hat{j} + \beta\hat{k}b=i^+3j^​+βk^ and c⃗=−i^+2j^−3k^\vec{c} = -\hat{i} + 2\hat{j} - 3\hat{k}c=−i^+2j^​−3k^ be three vectors such that, ∣b⃗×c⃗∣=53\left|\vec{b} \times \vec{c}\right| = 5\sqrt{3}​b×c​=53​ and a⃗\vec{a}a is perpendicular to b⃗\vec{b}b. Then the greatest amongst the values of ∣a⃗∣2\left|\vec{a}\right|^{2}∣a∣2 is _______.

Correct answer: 90

Step-by-step solution →
Q172·MathematicsNumericalJEE Main 2021
If the projection of the vector i^+2j^+k^\hat{i} + 2\hat{j} + \hat{k}i^+2j^​+k^ on the sum of the two vectors 2i^+4j^−5k^2\hat{i} + 4\hat{j} - 5\hat{k}2i^+4j^​−5k^ and −λi^+2j^+3k^-\lambda\hat{i} + 2\hat{j} + 3\hat{k}−λi^+2j^​+3k^ is 1, then λ\lambdaλ is equal to ________.

Correct answer: 5

Step-by-step solution →
Q173·MathematicsSingle correctJEE Main 2021
Let a⃗=i^+j^+k^\vec{a} = \hat{i} + \hat{j} + \hat{k}a=i^+j^​+k^ and b⃗=j^−k^\vec{b} = \hat{j} - \hat{k}b=j^​−k^. If c⃗\vec{c}c is a vector such that a⃗×c⃗=b⃗\vec{a} \times \vec{c} = \vec{b}a×c=b and a⃗⋅c⃗=3\vec{a} \cdot \vec{c} = 3a⋅c=3, then a⃗⋅(b⃗×c⃗)\vec{a} \cdot (\vec{b} \times \vec{c})a⋅(b×c) is equal to :
  1. (A)−2-2−2
  2. (B)−6-6−6
  3. (C)666
  4. (D)222

Correct answer: (A)

Step-by-step solution →
Q174·MathematicsNumericalJEE Main 2021
Let a⃗=i^+j^+k^\vec{a}=\hat{i}+\hat{j}+\hat{k}a=i^+j^​+k^, b⃗\vec{b}b and c⃗=j^−k^\vec{c}=\hat{j}-\hat{k}c=j^​−k^ be three vectors such that a⃗×b⃗=c⃗\vec{a} \times \vec{b}=\vec{c}a×b=c and a⃗⋅b⃗=1\vec{a} \cdot \vec{b}=1a⋅b=1. If the length of projection vector of the vector b⃗\vec{b}b on the vector a⃗×c⃗\vec{a} \times \vec{c}a×c is ℓ\ellℓ, then the value of 3ℓ23\ell^23ℓ2 is equal to ____.

Correct answer: 2

Step-by-step solution →
Q175·MathematicsSingle correctJEE Main 2021
Let a⃗,b⃗\vec{a}, \vec{b}a,b and c⃗\vec{c}c be three vectors such that a⃗=b⃗×(b⃗×c⃗)\vec{a} = \vec{b} \times \left( \vec{b} \times \vec{c} \right)a=b×(b×c). If magnitudes of the vectors a⃗,b⃗\vec{a}, \vec{b}a,b and c⃗\vec{c}c are √2, 1 and 2 respectively and the angle between b⃗\vec{b}b and c⃗\vec{c}c is θ(0<θ<π2)\theta\left( 0 < \theta < \frac{\pi}{2} \right)θ(0<θ<2π​), then the value of 1 + tan θ is equal to :
  1. (A)√3 + 1
  2. (B)3+13\frac{\sqrt{3} + 1}{\sqrt{3}}3​3​+1​
  3. (C)2
  4. (D)1

Correct answer: (C)

Step-by-step solution →
Q176·MathematicsNumericalJEE Main 2021
Let a⃗\vec{a}a = î − αĵ + βk̂, b⃗\vec{b}b = 3î + βĵ − αk̂ and c⃗\vec{c}c = −αî − 2ĵ + k̂, where α and β are integers. If a⃗⋅b⃗\vec{a}\cdot\vec{b}a⋅b = −1 and b⃗⋅c⃗\vec{b}\cdot\vec{c}b⋅c = 10, then (a⃗×b⃗)⋅c⃗\left(\vec{a}\times\vec{b}\right)\cdot\vec{c}(a×b)⋅c is equal to…….

Correct answer: 9

Step-by-step solution →
Q177·MathematicsSingle correctJEE Main 2021
Let a⃗=i^+j^+2k^\vec{a} = \hat{i} + \hat{j} + 2\hat{k}a=i^+j^​+2k^ and b⃗=−i^+2j^+3k^\vec{b} = -\hat{i} + 2\hat{j} + 3\hat{k}b=−i^+2j^​+3k^. Then the vector product (a⃗+b⃗)×(a⃗×((a⃗−b⃗)×b⃗))×b⃗\left(\vec{a} + \vec{b}\right) \times \left(\vec{a} \times \left(\left(\vec{a} - \vec{b}\right) \times \vec{b}\right)\right) \times \vec{b}(a+b)×(a×((a−b)×b))×b is equal to :
  1. (A)5(34i^−5j^+3k^)5\left(34\hat{i} - 5\hat{j} + 3\hat{k}\right)5(34i^−5j^​+3k^)
  2. (B)5(30i^−5j^+7k^)5\left(30\hat{i} - 5\hat{j} + 7\hat{k}\right)5(30i^−5j^​+7k^)
  3. (C)7(30i^−5j^+7k^)7\left(30\hat{i} - 5\hat{j} + 7\hat{k}\right)7(30i^−5j^​+7k^)
  4. (D)7(34i^−5j^+3k^)7\left(34\hat{i} - 5\hat{j} + 3\hat{k}\right)7(34i^−5j^​+3k^)

Correct answer: (D)

Step-by-step solution →
Q178·MathematicsSingle correctJEE Main 2021
Let the vectors (2+a+b)i^+(a+2b+c)j^−(b+c)k^,\left(2+a+b\right)\hat{i}+\left(a+2b+c\right)\hat{j}-\left(b+c\right)\hat{k},(2+a+b)i^+(a+2b+c)j^​−(b+c)k^, (1+b)i^+2bj^−bk^\left(1+b\right)\hat{i}+2b\hat{j}-b\hat{k}(1+b)i^+2bj^​−bk^ and (2+b)i^+2bj^+(1−b)k^,a,b,c∈R\left(2+b\right)\hat{i}+2b\hat{j}+\left(1-b\right)\hat{k}, a,b,c \in R(2+b)i^+2bj^​+(1−b)k^,a,b,c∈R be co-planar. Then which of the following is true?
  1. (A)2a=b+c2a=b+c2a=b+c
  2. (B)2b=a+c2b=a+c2b=a+c
  3. (C)a=b+2ca=b+2ca=b+2c
  4. (D)3c=a+b3c=a+b3c=a+b

Correct answer: (B)

Step-by-step solution →
Q179·MathematicsSingle correctJEE Main 2021
If ∣a⃗∣=2,∣b⃗∣=5\left|\vec{a}\right| = 2, \left|\vec{b}\right| = 5∣a∣=2,​b​=5 and ∣a⃗×b⃗∣=8\left|\vec{a} \times \vec{b}\right| = 8​a×b​=8, then ∣a⃗.b⃗∣\left|\vec{a} . \vec{b}\right|​a.b​ is equal to :
  1. (A)5
  2. (B)4
  3. (C)6
  4. (D)3

Correct answer: (C)

Step-by-step solution →
Q180·MathematicsSingle correctJEE Main 2021
Let a, b and c be distinct positive numbers. If the vectors ai^+aj^+ck^,i^+k^a\hat{i}+a\hat{j}+c\hat{k},\hat{i}+\hat{k}ai^+aj^​+ck^,i^+k^ and ci^+cj^+bk^c\hat{i}+c\hat{j}+b\hat{k}ci^+cj^​+bk^ are co-planar, then c is equal to :
  1. (A)21a+1b\frac{2}{\frac{1}{a}+\frac{1}{b}}a1​+b1​2​
  2. (B)a+b2\frac{a+b}{2}2a+b​
  3. (C)1a+1b\frac{1}{a}+\frac{1}{b}a1​+b1​
  4. (D)ab\sqrt{ab}ab​

Correct answer: (D)

Step-by-step solution →
Q181·MathematicsNumericalJEE Main 2021
Let p⃗=2i^+3j^+k^\vec{p} = 2\hat{i} + 3\hat{j} + \hat{k}p​=2i^+3j^​+k^ and q⃗=i^+2j^+k^\vec{q} = \hat{i} + 2\hat{j} + \hat{k}q​=i^+2j^​+k^ be two vectors. If a vector : r⃗=(αi^+βj^+γk^)\vec{r} = \left( \alpha\hat{i} + \beta\hat{j} + \gamma\hat{k} \right)r=(αi^+βj^​+γk^) is perpendicular to each of the vectors (p⃗+q⃗)\left( \vec{p} + \vec{q} \right)(p​+q​) and (p⃗−q⃗)\left( \vec{p} - \vec{q} \right)(p​−q​), and ∣r⃗∣=3\left| \vec{r} \right| = \sqrt{3}∣r∣=3​, then ∣α∣+∣β∣+∣γ∣\left| \alpha \right| + \left| \beta \right| + \left| \gamma \right|∣α∣+∣β∣+∣γ∣ is equal to......

Correct answer: 3

Step-by-step solution →
Q182·MathematicsNumericalJEE Main 2021
If (a⃗+3b⃗)\left(\vec{a} + 3\vec{b}\right)(a+3b) is perpendicular to (7a⃗−5b⃗)\left(7\vec{a} - 5\vec{b}\right)(7a−5b) and (a⃗−4b⃗)\left(\vec{a} - 4\vec{b}\right)(a−4b) is perpendicular to (7a⃗−2b⃗)\left(7\vec{a} - 2\vec{b}\right)(7a−2b) , then the angle between a⃗\vec{a}a and b⃗\vec{b}b (in degrees) is.......

Correct answer: 60

Step-by-step solution →
Q183·MathematicsSingle correctJEE Main 2021
Let a vector a⃗\vec{a}a be coplanar with vectors b⃗=2i^+j^+k^\vec{b} = 2\hat{i} + \hat{j} + \hat{k}b=2i^+j^​+k^ and c⃗=i^−j^+k^\vec{c} = \hat{i} - \hat{j} + \hat{k}c=i^−j^​+k^. If a⃗\vec{a}a is perpendicular to d⃗=3i^+2j^+6k^\vec{d} = 3\hat{i} + 2\hat{j} + 6\hat{k}d=3i^+2j^​+6k^ and ∣a⃗∣=10\left|\vec{a}\right| = \sqrt{10}∣a∣=10​. Then a possible value of [a⃗b⃗c⃗]+[a⃗b⃗d⃗]+[a⃗c⃗d⃗]\left[\vec{a}\vec{b}\vec{c}\right] + \left[\vec{a}\vec{b}\vec{d}\right] + \left[\vec{a}\vec{c}\vec{d}\right][abc]+[abd]+[acd] is equal to :
  1. (A)−40-40−40
  2. (B)−38-38−38
  3. (C)−42-42−42
  4. (D)−29-29−29

Correct answer: (C)

Step-by-step solution →
Q184·MathematicsSingle correctJEE Main 2021
Let three vectors a⃗,b⃗\vec{a}, \vec{b}a,b and c⃗\vec{c}c be such that a⃗×b⃗=c⃗\vec{a} \times \vec{b} = \vec{c}a×b=c, b⃗×c⃗=a⃗\vec{b} \times \vec{c} = \vec{a}b×c=a and ∣a⃗∣=2\left|\vec{a}\right| = 2∣a∣=2. Then which one of the following is not true?
  1. (A)Projection of a⃗\vec{a}a on (b⃗×c⃗)\left(\vec{b} \times \vec{c}\right)(b×c) is 2
  2. (B)[a⃗b⃗c⃗]+[c⃗a⃗b⃗]=8\left[\vec{a}\vec{b}\vec{c}\right] + \left[\vec{c}\vec{a}\vec{b}\right] = 8[abc]+[cab]=8
  3. (C)∣3a⃗+b⃗−2c⃗∣2=51\left|3\vec{a} + \vec{b} - 2\vec{c}\right|^2 = 51​3a+b−2c​2=51
  4. (D)a⃗×((b⃗+c⃗)×(b⃗−c⃗))=0⃗\vec{a} \times \left(\left(\vec{b} + \vec{c}\right) \times \left(\vec{b} - \vec{c}\right)\right) = \vec{0}a×((b+c)×(b−c))=0

Correct answer: (C)

Step-by-step solution →
Q185·MathematicsSingle correctJEE Main 2021
Let a⃗=2i^+j^−2k^\vec{a} = 2\hat{i}+\hat{j}-2\hat{k}a=2i^+j^​−2k^ and b⃗=i^+j^\vec{b} = \hat{i}+\hat{j}b=i^+j^​. If c⃗\vec{c}c is a vector such that a⃗.c⃗=∣c⃗∣\vec{a}.\vec{c} = |\vec{c}|a.c=∣c∣, ∣c⃗−a⃗∣=22\left|\vec{c}-\vec{a}\right| = 2\sqrt{2}∣c−a∣=22​ and the angle between (a⃗×b⃗)\left(\vec{a} \times \vec{b}\right)(a×b) and c⃗\vec{c}c is π6\frac{\pi}{6}6π​, then the value of ∣(a⃗×b⃗)×c⃗∣\left|\left(\vec{a} \times \vec{b}\right) \times \vec{c}\right|​(a×b)×c​ is :
  1. (A)3
  2. (B)23\frac{2}{3}32​
  3. (C)32\frac{3}{2}23​
  4. (D)4

Correct answer: (C)

Step-by-step solution →
Q186·MathematicsNumericalJEE Main 2021
For p>0p>0p>0, a vector v⃗2=2i^+(p+1)j^\vec{v}_{2}=2\hat{i}+(p+1)\hat{j}v2​=2i^+(p+1)j^​ is obtained by rotating the vector v⃗1=3 pi^+j^\vec{v}_{1}=\sqrt{3}\,p\hat{i}+\hat{j}v1​=3​pi^+j^​ by an angle θ\thetaθ about origin in counter clockwise direction. If tan⁡θ=(α3−2)(43+3)\tan\theta=\frac{\left(\alpha\sqrt{3}-2\right)}{\left(4\sqrt{3}+3\right)}tanθ=(43​+3)(α3​−2)​, then the value of α\alphaα is equal to……….

Correct answer: 6

Step-by-step solution →
Q187·MathematicsSingle correctJEE Main 2021
In a triangle ABC, if ∣BC→∣=3,∣CA→∣=5\left|\overrightarrow{BC}\right|=3,\left|\overrightarrow{CA}\right|=5​BC​=3,​CA​=5 and ∣BA→∣=7,\left|\overrightarrow{BA}\right|=7,​BA​=7, then the projection of the vector BA→\overrightarrow{BA}BA on BC→\overrightarrow{BC}BC is equal to :
  1. (A)192\frac{19}{2}219​
  2. (B)132\frac{13}{2}213​
  3. (C)152\frac{15}{2}215​
  4. (D)112\frac{11}{2}211​

Correct answer: (D)

Step-by-step solution →
Q188·MathematicsNumericalJEE Main 2021
Let a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c}a,b,c be three mutually perpendicular vectors of the same magnitude and equally inclined at an angle θ\thetaθ, with the vector a⃗+b⃗+c⃗\vec{a} + \vec{b} + \vec{c}a+b+c. Then 36cos⁡22θ36\cos^2 2\theta36cos22θ is equal to.......

Correct answer: 4

Step-by-step solution →
Q189·MathematicsSingle correctJEE Main 2021
In a triangle ABC, if ∣BC⃗∣=8\left|\vec{BC}\right| = 8​BC​=8, ∣CA⃗∣=7\left|\vec{CA}\right| = 7​CA​=7, ∣AB⃗∣=10\left|\vec{AB}\right| = 10​AB​=10, then the projection of the vector AB⃗\vec{AB}AB on AC⃗\vec{AC}AC is equal to :
  1. (A)254\frac{25}{4}425​
  2. (B)8514\frac{85}{14}1485​
  3. (C)12720\frac{127}{20}20127​
  4. (D)11516\frac{115}{16}16115​

Correct answer: (B)

Step-by-step solution →
Q190·MathematicsSingle correctJEE Main 2021
A vector a⃗\vec{a}a has components 3p3p3p and 111 with respect to a rectangular cartesian system. This system is rotated through a certain angle about the origin in the counter clockwise sense. If, with respect to new system, a⃗\vec{a}a has components p+1p + 1p+1 and 10\sqrt{10}10​, then a value of p is equal to:
  1. (A)111
  2. (B)−54-\dfrac{5}{4}−45​
  3. (C)45\dfrac{4}{5}54​
  4. (D)−1-1−1

Correct answer: (D)

Step-by-step solution →
Q191·MathematicsSingle correctJEE Main 2021
Let a⃗\vec{a}a and b⃗\vec{b}b be two non-zero vectors perpendicular to each other and ∣a⃗∣=∣b⃗∣\left|\vec{a}\right| = \left|\vec{b}\right|∣a∣=​b​. If ∣a⃗×b⃗∣=∣a⃗∣\left|\vec{a} \times \vec{b}\right| = \left|\vec{a}\right|​a×b​=∣a∣, then the angle between the vectors (a⃗+b⃗+(a⃗×b⃗))\left(\vec{a} + \vec{b} + \left(\vec{a} \times \vec{b}\right)\right)(a+b+(a×b)) and a⃗\vec{a}a is equal to :
  1. (A)sin⁡−1(13)\sin^{-1}\left(\frac{1}{\sqrt{3}}\right)sin−1(3​1​)
  2. (B)cos⁡−1(13)\cos^{-1}\left(\frac{1}{\sqrt{3}}\right)cos−1(3​1​)
  3. (C)cos⁡−1(12)\cos^{-1}\left(\frac{1}{\sqrt{2}}\right)cos−1(2​1​)
  4. (D)sin⁡−1(16)\sin^{-1}\left(\frac{1}{\sqrt{6}}\right)sin−1(6​1​)

Correct answer: (B)

Step-by-step solution →
Q192·MathematicsSingle correctJEE Main 2021
Let O be the origin. Let OP→=xi^+yj^−k^\overrightarrow{OP} = x\hat{i}+y\hat{j}-\hat{k}OP=xi^+yj^​−k^ and OQ→=−i^+2j^+3xk^\overrightarrow{OQ} = -\hat{i}+2\hat{j}+3x\hat{k}OQ​=−i^+2j^​+3xk^, x,y∈Rx, y \in Rx,y∈R, x>0x > 0x>0, be such that ∣PQ→∣=20\left|\overrightarrow{PQ}\right| = \sqrt{20}​PQ​​=20​ and the vector OP→\overrightarrow{OP}OP is perpendicular to OQ→\overrightarrow{OQ}OQ​. If OR→=3i^+zj^−7k^\overrightarrow{OR} = 3\hat{i}+z\hat{j}-7\hat{k}OR=3i^+zj^​−7k^, z∈Rz \in Rz∈R, is coplanar with OP→\overrightarrow{OP}OP and OQ→\overrightarrow{OQ}OQ​, then the value of x2+y2+z2x^{2}+y^{2}+z^{2}x2+y2+z2 is equal to
  1. (A)777
  2. (B)999
  3. (C)222
  4. (D)111

Correct answer: (B)

Step-by-step solution →
Q193·MathematicsSingle correctJEE Main 2021
Let a⃗=2i^−3j^+4k^\vec{a} = 2\hat{i} - 3\hat{j} + 4\hat{k}a=2i^−3j^​+4k^ and b⃗=7i^+j^−6k^\vec{b} = 7\hat{i} + \hat{j} - 6\hat{k}b=7i^+j^​−6k^. If r⃗×a⃗=r⃗×b⃗,r⃗⋅(i^+2j^+k^)=−3\vec{r} \times \vec{a} = \vec{r} \times \vec{b}, \vec{r} \cdot \left(\hat{i} + 2\hat{j} + \hat{k}\right) = -3r×a=r×b,r⋅(i^+2j^​+k^)=−3, then r⃗⋅(2i^−3j^+k^)\vec{r} \cdot \left(2\hat{i} - 3\hat{j} + \hat{k}\right)r⋅(2i^−3j^​+k^) is equal to :
  1. (A)12
  2. (B)8
  3. (C)13
  4. (D)10

Correct answer: (A)

Step-by-step solution →
Q194·MathematicsNumericalJEE Main 2021
Let x⃗\vec{x}x be a vector in the plane containing vectors a⃗=2i^−j^+k^\vec{a} = 2\hat{i}-\hat{j}+\hat{k}a=2i^−j^​+k^ and b⃗=i^+2j^−k^\vec{b} = \hat{i}+2\hat{j}-\hat{k}b=i^+2j^​−k^. If the vector x⃗\vec{x}x is perpendicular to (3i^+2j^−k^)\left(3\hat{i}+2\hat{j}-\hat{k}\right)(3i^+2j^​−k^) and its projection on a⃗\vec{a}a is 1762\frac{17\sqrt{6}}{2}2176​​, then the value of ∣x⃗∣2\left|\vec{x}\right|^{2}∣x∣2 is equal to ________.

Correct answer: 486

Step-by-step solution →
Q195·MathematicsNumericalJEE Main 2021
If a⃗=αi^+βj^+3k^\vec{a} = \alpha\hat{i} + \beta\hat{j} + 3\hat{k}a=αi^+βj^​+3k^, b⃗=−βi^−αj^−k^\vec{b} = -\beta\hat{i} - \alpha\hat{j} - \hat{k}b=−βi^−αj^​−k^ and c⃗=i^−2j^−k^\vec{c} = \hat{i} - 2\hat{j} - \hat{k}c=i^−2j^​−k^ such that a⃗⋅b⃗=1\vec{a} \cdot \vec{b} = 1a⋅b=1 and b⃗⋅c⃗=−3\vec{b} \cdot \vec{c} = -3b⋅c=−3, then 13((a⃗×b⃗)⋅c⃗)\frac{1}{3}\left(\left(\vec{a} \times \vec{b}\right) \cdot \vec{c}\right)31​((a×b)⋅c) is equal to _______.

Correct answer: 2

Step-by-step solution →
Q196·MathematicsSingle correctJEE Main 2021
Let a⃗=i^+2j^−3k^\vec{a}=\hat{i}+2\hat{j}-3\hat{k}a=i^+2j^​−3k^ and b⃗=2i^−3j^+5k^\vec{b}=2\hat{i}-3\hat{j}+5\hat{k}b=2i^−3j^​+5k^. If r⃗×a⃗=b⃗×r⃗\vec{r}\times\vec{a}=\vec{b}\times\vec{r}r×a=b×r, r⃗.(αi^+2j^+k^)=3\vec{r}.\left(\alpha\hat{i}+2\hat{j}+\hat{k}\right)=3r.(αi^+2j^​+k^)=3 and r⃗.(2i^+5j^−αk^)=−1\vec{r}.\left(2\hat{i}+5\hat{j}-\alpha\hat{k}\right)=-1r.(2i^+5j^​−αk^)=−1, α∈R\alpha \in Rα∈R, then the value of α+∣r⃗∣2\alpha+\left|\vec{r}\right|^{2}α+∣r∣2 is equal to :
  1. (A)999
  2. (B)151515
  3. (C)131313
  4. (D)111111

Correct answer: (B)

Step-by-step solution →
Q197·MathematicsSingle correctJEE Main 2021
Let a vector αi^+βj^\alpha\hat{i} + \beta\hat{j}αi^+βj^​ be obtained by rotating the vector 3i^+j^\sqrt{3}\hat{i} + \hat{j}3​i^+j^​ by an angle 45° about the origin in counterclockwise direction in the first quadrant. Then the area of triangle having vertices (α, β), (0, β) and (0, 0) is equal to
  1. (A)12\frac{1}{2}21​
  2. (B)1
  3. (C)12\frac{1}{\sqrt{2}}2​1​
  4. (D)222\sqrt{2}22​

Correct answer: (A)

Step-by-step solution →
Q198·MathematicsNumericalJEE Main 2021
Let c⃗\vec{c}c be a vector perpendicular to the vectors a⃗=i^+j^−k^\vec{a}=\hat{i}+\hat{j}-\hat{k}a=i^+j^​−k^ and b⃗=i^+2j^+k^\vec{b}=\hat{i}+2\hat{j}+\hat{k}b=i^+2j^​+k^. If c⃗.(i^+j^+3k^)=8\vec{c}.\left(\hat{i}+\hat{j}+3\hat{k}\right)=8c.(i^+j^​+3k^)=8 then the value of c⃗.(a⃗×b⃗)\vec{c}.\left(\vec{a}\times\vec{b}\right)c.(a×b) is equal to _________.

Correct answer: 28

Step-by-step solution →
Q199·MathematicsSingle correctJEE Main 2021
Let the position vectors of two points P and Q be 3i^−j^+2k^3\hat{i} - \hat{j} + 2\hat{k}3i^−j^​+2k^ and i^+2j^−4k^\hat{i} + 2\hat{j} - 4\hat{k}i^+2j^​−4k^, respectively. Let R and S be two points such that the direction ratios of lines PR and QS are (4, –1, 2) and (–2, 1, –2), respectively. Let lines PR and QS intersect at T. If the vector TA→\overrightarrow{TA}TA is perpendicular to both PR→\overrightarrow{PR}PR and QS→\overrightarrow{QS}QS​ and the length of vector TA→\overrightarrow{TA}TA is 5\sqrt{5}5​ units, then the modulus of a position vector of A is :
  1. (A)482\sqrt{482}482​
  2. (B)171\sqrt{171}171​
  3. (C)5\sqrt{5}5​
  4. (D)227\sqrt{227}227​

Correct answer: (B)

Step-by-step solution →
Q200·MathematicsSingle correctJEE Main 2021
If vectors a⃗1=xi^−j^+k^\vec{a}_{1} = x\hat{i} - \hat{j} + \hat{k}a1​=xi^−j^​+k^ and a⃗2=i^+yj^+zk^\vec{a}_{2} = \hat{i} + y\hat{j} + z\hat{k}a2​=i^+yj^​+zk^ are collinear, then a possible unit vector parallel to the vector xi^+yj^+zk^x\hat{i} + y\hat{j} + z\hat{k}xi^+yj^​+zk^ is :
  1. (A)12(−j^+k^)\frac{1}{\sqrt{2}} (-\hat{j} + \hat{k})2​1​(−j^​+k^)
  2. (B)12(i^−j^)\frac{1}{\sqrt{2}} (\hat{i} - \hat{j})2​1​(i^−j^​)
  3. (C)13(i^−j^+k^)\frac{1}{\sqrt{3}} (\hat{i} - \hat{j} + \hat{k})3​1​(i^−j^​+k^)
  4. (D)13(i^+j^−k^)\frac{1}{\sqrt{3}} (\hat{i} + \hat{j} - \hat{k})3​1​(i^+j^​−k^)

Correct answer: (C)

Step-by-step solution →
Q201·MathematicsSingle correctJEE Main 2021
If a⃗\vec{a}a & b⃗\vec{b}b are perpendicular vactors, then a⃗×(a⃗×(a⃗×(a⃗×b⃗)))\vec{a} \times \left( \vec{a} \times \left( \vec{a} \times \left( \vec{a} \times \vec{b} \right) \right) \right)a×(a×(a×(a×b))) is equal to
  1. (A)12∣a⃗∣4 b⃗\frac{1}{2}\left|\vec{a}\right|^{4}\,\vec{b}21​∣a∣4b
  2. (B)a⃗×b⃗\vec{a} \times \vec{b}a×b
  3. (C)∣a⃗∣4 b⃗\left|\vec{a}\right|^{4}\,\vec{b}∣a∣4b
  4. (D)0⃗\vec{0}0

Correct answer: (C)

Step-by-step solution →
Q202·MathematicsNumericalJEE Main 2021
Let a=ˆi+α+ˆj3kˆ andb=3iˆ−α+ˆjkˆ . If the area of the parallelogram whose adjacent sides are represented by the vectors a and b is 8 3 square units, then a·b is equal to ______.

Correct answer: 2

Step-by-step solution →
Q203·MathematicsNumericalJEE Main 2021
Let a⃗=i^+2j^−k^\vec{a} = \hat{i} + 2\hat{j} - \hat{k}a=i^+2j^​−k^, b⃗=i^−j^\vec{b} = \hat{i} - \hat{j}b=i^−j^​ and c⃗=i^−j^−k^\vec{c} = \hat{i} - \hat{j} - \hat{k}c=i^−j^​−k^ be three given vectors. If r⃗\vec{r}r is a vector such that r⃗×a⃗=c⃗×a⃗\vec{r} \times \vec{a} = \vec{c} \times \vec{a}r×a=c×a and r⃗⋅b⃗=0\vec{r} \cdot \vec{b} = 0r⋅b=0, then r⃗⋅a⃗\vec{r} \cdot \vec{a}r⋅a is equal to ________

Correct answer: 12

Step-by-step solution →
Q204·MathematicsNumericalJEE Main 2021
Let three vectors a⃗,b⃗\vec{a}, \vec{b}a,b and c⃗\vec{c}c be such that c⃗\vec{c}c is coplanar with a⃗\vec{a}a and b⃗\vec{b}b, a⃗⋅c⃗=7\vec{a} \cdot \vec{c} = 7a⋅c=7 and b⃗\vec{b}b is perpendicular to c⃗\vec{c}c, where a⃗=−i^+j^+k^\vec{a} = -\hat{i} + \hat{j} + \hat{k}a=−i^+j^​+k^ and b⃗=2i^+k^\vec{b} = 2\hat{i} + \hat{k}b=2i^+k^, then the value of 2∣a⃗+b⃗+c⃗∣22\left|\vec{a} + \vec{b} + \vec{c}\right|^22​a+b+c​2 is ________

Correct answer: 75

Step-by-step solution →
Q205·MathematicsMultiple correctJEE Advanced 2020
Let aaa and bbb be positive real numbers. Suppose PQ→=ai^+bj^\overrightarrow{PQ} = a\hat{i} + b\hat{j}PQ​=ai^+bj^​ and PS→=ai^−bj^\overrightarrow{PS} = a\hat{i} - b\hat{j}PS=ai^−bj^​ are adjacent sides of a parallelogram PQRSPQRSPQRS. Let u⃗\vec{u}u and v⃗\vec{v}v be the projection vectors of w⃗=i^+j^\vec{w} = \hat{i} + \hat{j}w=i^+j^​ along PQ→\overrightarrow{PQ}PQ​ and PS→\overrightarrow{PS}PS, respectively. If ∣u⃗∣+∣v⃗∣=∣w⃗∣|\vec{u}| + |\vec{v}| = |\vec{w}|∣u∣+∣v∣=∣w∣ and if the area of the parallelogram PQRSPQRSPQRS is 8, then which of the following statements is/are TRUE?
  1. (A)a+b=4a + b = 4a+b=4
  2. (B)a−b=2a - b = 2a−b=2
  3. (C)The length of the diagonal PRPRPR of the parallelogram PQRSPQRSPQRS is 4
  4. (D)w⃗\vec{w}w is an angle bisector of the vectors PQ→\overrightarrow{PQ}PQ​ and PS→\overrightarrow{PS}PS

Correct answer: (A), (C)

Step-by-step solution →
Q206·MathematicsNumericalJEE Advanced 2020
In a triangle PQR, let a⃗=QR→\vec{a} = \overrightarrow{QR}a=QR​, b⃗=RP→\vec{b} = \overrightarrow{RP}b=RP and c⃗=PQ→\vec{c} = \overrightarrow{PQ}c=PQ​. If ∣a⃗∣=3|\vec{a}| = 3∣a∣=3, ∣b⃗∣=4|\vec{b}| = 4∣b∣=4 and a⃗⋅(c⃗−b⃗)c⃗⋅(a⃗−b⃗)=∣a⃗∣∣a⃗∣+∣b⃗∣\frac{\vec{a}\cdot(\vec{c} - \vec{b})}{\vec{c}\cdot(\vec{a} - \vec{b})} = \frac{|\vec{a}|}{|\vec{a}| + |\vec{b}|}c⋅(a−b)a⋅(c−b)​=∣a∣+∣b∣∣a∣​, then the value of ∣a⃗×b⃗∣2\left|\vec{a} \times \vec{b}\right|^{2}​a×b​2 is ______

Correct answer: 108.00

Step-by-step solution →
Q207·MathematicsNumericalJEE Main 2020
If a⃗\vec{a}a and b⃗\vec{b}b are unit vectors, then the greatest value of 3∣a⃗+b⃗∣+∣a⃗−b⃗∣\sqrt{3}\left|\vec{a}+\vec{b}\right|+\left|\vec{a}-\vec{b}\right|3​​a+b​+​a−b​ is ____.

Correct answer: 04

Step-by-step solution →
Q208·MathematicsNumericalJEE Main 2020
If x⃗\vec{x}x and y⃗\vec{y}y​ be two non-zero vectors such that ∣x⃗+y⃗∣=∣x⃗∣|\vec{x} + \vec{y}| = |\vec{x}|∣x+y​∣=∣x∣ and 2x⃗+λy⃗2\vec{x} + \lambda\vec{y}2x+λy​ is perpendicular to y⃗\vec{y}y​, then the value of λ\lambdaλ is

Correct answer: 1

Step-by-step solution →
Q209·MathematicsSingle correctJEE Main 2020
If the volume of a parallelepiped, whose coterminus edges are given by the vectors a⃗=i^+j^+nk^,b⃗=2i^+4j^−nk^\vec{a}=\hat{i}+\hat{j}+n\hat{k}, \vec{b}=2\hat{i}+4\hat{j}-n\hat{k}a=i^+j^​+nk^,b=2i^+4j^​−nk^ and c⃗=i^+nj^+3k^\vec{c}=\hat{i}+n\hat{j}+3\hat{k}c=i^+nj^​+3k^ (n≥0)(n \ge 0)(n≥0), is 158 cu.units. then:
  1. (A)a⃗.c⃗=14\vec{a}.\vec{c}=14a.c=14
  2. (B)n=7
  3. (C)b⃗.c⃗=10\vec{b}.\vec{c}=10b.c=10
  4. (D)n=9

Correct answer: (C)

Step-by-step solution →
Q210·MathematicsNumericalJEE Main 2020
Let the vectors a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c}a,b,c be such that ∣a⃗∣=2,∣b⃗∣=4|\vec{a}| = 2, |\vec{b}| = 4∣a∣=2,∣b∣=4 and ∣c⃗∣=4|\vec{c}| = 4∣c∣=4. if the projection of b⃗\vec{b}b on a⃗\vec{a}a is equal to the projection of c⃗\vec{c}c on a⃗\vec{a}a and b⃗\vec{b}b is perpendicular to c⃗\vec{c}c, then the value of ∣a⃗+b⃗−c⃗∣|\vec{a} + \vec{b} - \vec{c}|∣a+b−c∣ is __________.

Correct answer: 6.00

Step-by-step solution →
Q211·MathematicsSingle correctJEE Main 2020
Let x0x_{0}x0​ be the point of local maxima of f(x)=a⃗⋅(b⃗×c⃗)f(x) = \vec{a} \cdot \left( \vec{b} \times \vec{c} \right)f(x)=a⋅(b×c) where a⃗=xi^−2j^+3k^\vec{a} = x\hat{i} - 2\hat{j} + 3\hat{k}a=xi^−2j^​+3k^, b⃗=−2i^+xj^−k^\vec{b} = -2\hat{i} + x\hat{j} - \hat{k}b=−2i^+xj^​−k^ and c⃗=7i^−2j^+xk^\vec{c} = 7\hat{i} - 2\hat{j} + x\hat{k}c=7i^−2j^​+xk^. Then the value of a⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}a⋅b+b⋅c+c⋅a at x=x0x = x_{0}x=x0​ is
  1. (A)−4-4−4
  2. (B)−22-22−22
  3. (C)−30-30−30
  4. (D)141414

Correct answer: (B)

Step-by-step solution →
Q212·MathematicsNumericalJEE Main 2020
If a⃗=2i^+j^+2k^\vec{a} = 2\hat{i} + \hat{j} + 2\hat{k}a=2i^+j^​+2k^, then the value of ∣i^×(a⃗×i^)∣2+∣j^×(a⃗×j^)∣2+∣k^×(a⃗×k^)∣2\left|\hat{i} \times \left(\vec{a} \times \hat{i}\right)\right|^{2} + \left|\hat{j} \times \left(\vec{a} \times \hat{j}\right)\right|^{2} + \left|\hat{k} \times \left(\vec{a} \times \hat{k}\right)\right|^{2}​i^×(a×i^)​2+​j^​×(a×j^​)​2+​k^×(a×k^)​2 is equal to:

Correct answer: 18

Step-by-step solution →
Q213·MathematicsSingle correctJEE Main 2020
Let a,b,c∈Ra, b, c \in Ra,b,c∈R be such that a2+b2+c2=1a^2 + b^2 + c^2 = 1a2+b2+c2=1. If acos⁡θ=bcos⁡(θ+2π3)=ccos⁡(θ+4π3)a\cos\theta = b\cos\left(\theta + \frac{2\pi}{3}\right) = c\cos\left(\theta + \frac{4\pi}{3}\right)acosθ=bcos(θ+32π​)=ccos(θ+34π​), where θ=π9\theta = \frac{\pi}{9}θ=9π​, then the angle between the vectors ai^+bj^+ck^a\hat{i} + b\hat{j} + c\hat{k}ai^+bj^​+ck^ and bi^+cj^+ak^b\hat{i} + c\hat{j} + a\hat{k}bi^+cj^​+ak^ is:
  1. (A)π9\frac{\pi}{9}9π​
  2. (B)2π3\frac{2\pi}{3}32π​
  3. (C)000
  4. (D)π2\frac{\pi}{2}2π​

Correct answer: (D)

Step-by-step solution →
Q214·MathematicsNumericalJEE Main 2020
If the vectors, p⃗=(a+1)i^+aj^+ak^\vec{p}=(a+1)\hat{i}+a\hat{j}+a\hat{k}p​=(a+1)i^+aj^​+ak^, q⃗=ai^+(a+1)j^+ak^\vec{q}=a\hat{i}+(a+1)\hat{j}+a\hat{k}q​=ai^+(a+1)j^​+ak^ and r⃗=ai^+aj^+(a+1)k^\vec{r}=a\hat{i}+a\hat{j}+(a+1)\hat{k}r=ai^+aj^​+(a+1)k^ (a∈R)(a \in R)(a∈R) are coplanar and 3(p⃗.q⃗)2−λ∣r⃗×q⃗∣2=03(\vec{p}.\vec{q})^{2}-\lambda|\vec{r}\times\vec{q}|^{2}=03(p​.q​)2−λ∣r×q​∣2=0, then the value of λ\lambdaλ is ______.

Correct answer: 1

Step-by-step solution →
Q215·MathematicsNumericalJEE Main 2020
Let a⃗,b⃗\vec{a}, \vec{b}a,b and c⃗\vec{c}c be three vectors such that ∣a⃗∣=3,∣b⃗∣=5,b⃗.c⃗=10|\vec{a}| = \sqrt{3}, |\vec{b}| = 5, \vec{b}.\vec{c} = 10∣a∣=3​,∣b∣=5,b.c=10 and the angle between b⃗\vec{b}b and c⃗\vec{c}c is π3\frac{\pi}{3}3π​. If a⃗\vec{a}a is perpendicular to the vector b⃗×c⃗\vec{b} \times \vec{c}b×c, then ∣a⃗×(b⃗×c⃗)∣|\vec{a} \times (\vec{b} \times \vec{c})|∣a×(b×c)∣ is equal to ____.

Correct answer: 30

Step-by-step solution →
Q216·MathematicsSingle correctJEE Main 2020
Let a⃗=i^−2j^+k^\vec{a}=\hat{i}-2\hat{j}+\hat{k}a=i^−2j^​+k^ and b⃗=i^−j^+k^\vec{b}=\hat{i}-\hat{j}+\hat{k}b=i^−j^​+k^ be two vectors. If c⃗\vec{c}c is a vector such that b⃗×c⃗=b⃗×a⃗\vec{b}\times\vec{c}=\vec{b}\times\vec{a}b×c=b×a and c⃗.a⃗=0\vec{c}.\vec{a}=0c.a=0, then c⃗.b⃗\vec{c}.\vec{b}c.b is equal to:
  1. (A)−12-\dfrac{1}{2}−21​
  2. (B)−32-\dfrac{3}{2}−23​
  3. (C)−1-1−1
  4. (D)12\dfrac{1}{2}21​

Correct answer: (A)

Step-by-step solution →
Q217·MathematicsSingle correctJEE Main 2020
Let the volume of a parallelepiped whose coterminous edges are given by u⃗=i^+j^+λk^\vec{u}=\hat{i}+\hat{j}+\lambda\hat{k}u=i^+j^​+λk^, v⃗=i^+j^+3k^\vec{v}=\hat{i}+\hat{j}+3\hat{k}v=i^+j^​+3k^ and w⃗=2i^+j^+k^\vec{w}=2\hat{i}+\hat{j}+\hat{k}w=2i^+j^​+k^ be 1 cu. Unit. If θ\thetaθ be the angle between the edges u⃗\vec{u}u and w⃗\vec{w}w, then cos⁡θ\cos\thetacosθ can be:
  1. (A)763\dfrac{7}{6\sqrt{3}}63​7​
  2. (B)57\dfrac{5}{7}75​
  3. (C)533\dfrac{5}{3\sqrt{3}}33​5​
  4. (D)766\dfrac{7}{6\sqrt{6}}66​7​

Correct answer: (A)

Step-by-step solution →
Q218·MathematicsSingle correctJEE Main 2020
A vector a⃗=αi^+2j^+βk^ (α,β∈R)\vec{a} = \alpha\hat{i} + 2\hat{j} + \beta\hat{k} \ (\alpha,\beta \in R)a=αi^+2j^​+βk^ (α,β∈R) lies in the plane of the vectors, b⃗=i^+j^\vec{b} = \hat{i} + \hat{j}b=i^+j^​ and c⃗=i^−j^+4k^\vec{c} = \hat{i} - \hat{j} + 4\hat{k}c=i^−j^​+4k^. If a⃗\vec{a}a bisects the angle between b⃗\vec{b}b and c⃗\vec{c}c, then:
  1. (A)a⃗.k^+4=0\vec{a}.\hat{k} + 4 = 0a.k^+4=0
  2. (B)a⃗.k^+2=0\vec{a}.\hat{k} + 2 = 0a.k^+2=0
  3. (C)a⃗.i^+3=0\vec{a}.\hat{i} + 3 = 0a.i^+3=0
  4. (D)a⃗.i^+1=0\vec{a}.\hat{i} + 1 = 0a.i^+1=0

Correct answer: (B)

Step-by-step solution →
Q219·MathematicsNumericalJEE Advanced 2019
Let a⃗=2i^+j^−k^\vec{a} = 2\hat{i} + \hat{j} - \hat{k}a=2i^+j^​−k^ and b⃗=i^+2j^+k^\vec{b} = \hat{i} + 2\hat{j} + \hat{k}b=i^+2j^​+k^ be two vectors. Consider a vector c⃗=αa⃗+βb⃗\vec{c} = \alpha\vec{a} + \beta\vec{b}c=αa+βb, α, β∈R\alpha,\ \beta \in \mathbb{R}α, β∈R. If the projection of c⃗\vec{c}c on the vector (a⃗+b⃗)\left(\vec{a} + \vec{b}\right)(a+b) is 323\sqrt{2}32​, then the minimum value of (c⃗−(a⃗×b⃗))⋅c⃗\left(\vec{c} - \left(\vec{a} \times \vec{b}\right)\right)\cdot\vec{c}(c−(a×b))⋅c equals ____

Correct answer: 18.00

Step-by-step solution →
Q220·MathematicsSingle correctJEE Main 2019
Let a⃗=3i^+2j^+2k^\vec{a} = 3\hat{i} + 2\hat{j} + 2\hat{k}a=3i^+2j^​+2k^ and b⃗=i^+2j^−2k^\vec{b} = \hat{i} + 2\hat{j} - 2\hat{k}b=i^+2j^​−2k^ be two vectors. If a vector perpendicular to both the vectors a⃗+b⃗\vec{a} + \vec{b}a+b and a⃗−b⃗\vec{a} - \vec{b}a−b has the magnitude 12 then one such vector is:
  1. (A)4(2i^−2j^−k^)4\left(2\hat{i} - 2\hat{j} - \hat{k}\right)4(2i^−2j^​−k^)
  2. (B)4(2i^−2j^+k^)4\left(2\hat{i} - 2\hat{j} + \hat{k}\right)4(2i^−2j^​+k^)
  3. (C)4(2i^+2j^+k^)4\left(2\hat{i} + 2\hat{j} + \hat{k}\right)4(2i^+2j^​+k^)
  4. (D)4(2i^+2j^−k^)4\left(2\hat{i} + 2\hat{j} - \hat{k}\right)4(2i^+2j^​−k^)

Correct answer: (A)

Step-by-step solution →
Q221·MathematicsSingle correctJEE Main 2019
Let α∈R\alpha \in Rα∈R and the three vectors a⃗=αi^+j^+3k^\vec{a} = \alpha\hat{i} + \hat{j} + 3\hat{k}a=αi^+j^​+3k^, b⃗=2i^+j^−αk^\vec{b} = 2\hat{i} + \hat{j} - \alpha\hat{k}b=2i^+j^​−αk^ and c⃗=αi^−2j^+3k^\vec{c} = \alpha\hat{i} - 2\hat{j} + 3\hat{k}c=αi^−2j^​+3k^. Then the set S=(α:a⃗,b⃗ and c⃗ are coplanar)S = (\alpha : \vec{a}, \vec{b} \text{ and } \vec{c} \text{ are coplanar})S=(α:a,b and c are coplanar)
  1. (A)Contains exactly two numbers only one of which is positive
  2. (B)is empty
  3. (C)Contains exactly two positive numbers
  4. (D)is singleton

Correct answer: (B)

Step-by-step solution →
Q222·MathematicsSingle correctJEE Main 2019
Let A(3, 0, −1), B(2, 10, 6) and C(1, 2, 1) be the vertices of a triangle and M be the midpoint of AC. If G divides BM in the ratio, 2 : 1, then cos(∠\angle∠GOA) (O being he origin) is equal to:
  1. (A)130\dfrac{1}{\sqrt{30}}30​1​
  2. (B)1215\dfrac{1}{2\sqrt{15}}215​1​
  3. (C)1610\dfrac{1}{6\sqrt{10}}610​1​
  4. (D)115\dfrac{1}{\sqrt{15}}15​1​

Correct answer: (D)

Step-by-step solution →
Q223·MathematicsSingle correctJEE Main 2019
If a unit vector r⃗\vec{r}r makes angles π3\dfrac{\pi}{3}3π​ with i^\hat{i}i^, π4\dfrac{\pi}{4}4π​ with j^\hat{j}j^​ and θ∈(0,π)\theta\in\left(0,\pi\right)θ∈(0,π) with k^\hat{k}k^, then a value of θ\thetaθ is
  1. (A)5π12\dfrac{5\pi}{12}125π​
  2. (B)5π6\dfrac{5\pi}{6}65π​
  3. (C)2π3\dfrac{2\pi}{3}32π​
  4. (D)π4\dfrac{\pi}{4}4π​

Correct answer: (C)

Step-by-step solution →
Q224·MathematicsSingle correctJEE Main 2019
Let α⃗=3i^+j^\vec{\alpha}=3\hat{i}+\hat{j}α=3i^+j^​ and β⃗=2i^−j^+3k^\vec{\beta}=2\hat{i}-\hat{j}+3\hat{k}β​=2i^−j^​+3k^. If β⃗=β1⃗−β2⃗\vec{\beta}=\vec{\beta_{1}}-\vec{\beta_{2}}β​=β1​​−β2​​, where β1⃗\vec{\beta_{1}}β1​​ is parallel to α⃗\vec{\alpha}α and β2⃗\vec{\beta_{2}}β2​​ is perpendicular to α⃗\vec{\alpha}α, then β1⃗×β2⃗\vec{\beta_{1}}\times\vec{\beta_{2}}β1​​×β2​​ is equal to:
  1. (A)12(−3i^+9j^+5k^)\dfrac{1}{2}(-3\hat{i}+9\hat{j}+5\hat{k})21​(−3i^+9j^​+5k^)
  2. (B)12(3i^−9j^+5k^)\dfrac{1}{2}(3\hat{i}-9\hat{j}+5\hat{k})21​(3i^−9j^​+5k^)
  3. (C)−3i^+9j^+5k^-3\hat{i}+9\hat{j}+5\hat{k}−3i^+9j^​+5k^
  4. (D)3i^−9j^−5k^3\hat{i}-9\hat{j}-5\hat{k}3i^−9j^​−5k^

Correct answer: (A)

Step-by-step solution →
Q225·MathematicsSingle correctJEE Main 2019
Let a⃗=3i^+2j^+xk^\vec{a}=3\hat{i}+2\hat{j}+x\hat{k}a=3i^+2j^​+xk^ and b⃗=i^−−j^+k^\vec{b}=\hat{i}--\hat{j}+\hat{k}b=i^−−j^​+k^, for some real x. Then ∣a⃗×b⃗∣=r|\vec{a}\times\vec{b}|=r∣a×b∣=r is possible if:
  1. (A)r≥532r\geq5\sqrt{\frac{3}{2}}r≥523​​
  2. (B)332<r<5323\sqrt{\frac{3}{2}}<r<5\sqrt{\frac{3}{2}}323​​<r<523​​
  3. (C)32<r≤332\sqrt{\frac{3}{2}}<r\leq3\sqrt{\frac{3}{2}}23​​<r≤323​​
  4. (D)0<r≤320<r\leq\sqrt{\frac{3}{2}}0<r≤23​​

Correct answer: (A)

Step-by-step solution →
Q226·MathematicsSingle correctJEE Main 2019
The magnitude of the projection of the vector 2i^+3j^+k^2\hat{i}+3\hat{j}+\hat{k}2i^+3j^​+k^ on the vector perpendicular to the plane containing the vectors i^+j^+k^\hat{i}+\hat{j}+\hat{k}i^+j^​+k^ and i^+2j^+3k^\hat{i}+2\hat{j}+3\hat{k}i^+2j^​+3k^, is:
  1. (A)363\sqrt{6}36​
  2. (B)32\frac{\sqrt{3}}{2}23​​
  3. (C)6\sqrt{6}6​
  4. (D)32\sqrt{\frac{3}{2}}23​​

Correct answer: (D)

Step-by-step solution →
Q227·MathematicsSingle correctJEE Main 2019
The sum of the distinct real values of μ\muμ, for which the vectors, μi^+j^+k^\mu\hat{i} + \hat{j} + \hat{k}μi^+j^​+k^, i^+μj^+k^\hat{i} + \mu\hat{j} + \hat{k}i^+μj^​+k^, i^+j^+μk^\hat{i} + \hat{j} + \mu\hat{k}i^+j^​+μk^ are co-planar, is:
  1. (A)-1
  2. (B)0
  3. (C)1
  4. (D)2

Correct answer: (A)

Step-by-step solution →
Q228·MathematicsSingle correctJEE Main 2019
Let a⃗\vec{a}a, b⃗\vec{b}b, and c⃗\vec{c}c be three unit vectors, out of which vectors b⃗\vec{b}b and c⃗\vec{c}c are non-parallel. If α and β are the angles which vector a⃗\vec{a}a makes with vectors b⃗\vec{b}b and c⃗\vec{c}c respectively and a⃗×(b⃗×c⃗)=12b⃗\vec{a} \times (\vec{b} \times \vec{c}) = \dfrac{1}{2}\vec{b}a×(b×c)=21​b, then |α − β| is equal to :
  1. (A)30°30°30°
  2. (B)90°90°90°
  3. (C)60°60°60°
  4. (D)45°45°45°

Correct answer: (A)

Step-by-step solution →
Q229·MathematicsSingle correctJEE Main 2019
Let a⃗=i^+2j^+4k^,b⃗=i^+λj^+4k^\vec{a}=\hat{i}+2\hat{j}+4\hat{k}, \vec{b}=\hat{i}+\lambda\hat{j}+4\hat{k}a=i^+2j^​+4k^,b=i^+λj^​+4k^ and c⃗=2i^+4j^+(λ2−1)k^\vec{c}=2\hat{i}+4\hat{j}+\left(\lambda^{2}-1\right)\hat{k}c=2i^+4j^​+(λ2−1)k^ be coplanar vectors. Then the non − zero vector a⃗×c⃗\vec{a}\times\vec{c}a×c is:
  1. (A)−10i^−5j^-10\hat{i}-5\hat{j}−10i^−5j^​
  2. (B)−14i^−5j^-14\hat{i}-5\hat{j}−14i^−5j^​
  3. (C)−14i^+5j^-14\hat{i}+5\hat{j}−14i^+5j^​
  4. (D)−10i^+5j^-10\hat{i}+5\hat{j}−10i^+5j^​

Correct answer: (D)

Step-by-step solution →
Q230·MathematicsSingle correctJEE Main 2019
Let a⃗=2i^+λ1j^+3k^\vec{a} = 2\hat{i} + \lambda_1\hat{j} + 3\hat{k}a=2i^+λ1​j^​+3k^, b⃗=4i^+(3−λ2)j^+6k^\vec{b} = 4\hat{i} + (3 - \lambda_2)\hat{j} + 6\hat{k}b=4i^+(3−λ2​)j^​+6k^ and c⃗=3i^+6j^+(λ3−1)k^\vec{c} = 3\hat{i} + 6\hat{j} + (\lambda_3 - 1)\hat{k}c=3i^+6j^​+(λ3​−1)k^ be three vectors such that b⃗=2a⃗\vec{b} = 2\vec{a}b=2a and a⃗\vec{a}a is perpendicular to c⃗\vec{c}c. Then a possible value of (λ1,λ2,λ3)(\lambda_1, \lambda_2, \lambda_3)(λ1​,λ2​,λ3​) is:
  1. (A)(1, 3, 1)
  2. (B)(−12,4,0)\left(-\dfrac{1}{2}, 4, 0\right)(−21​,4,0)
  3. (C)(12,4,−2)\left(\dfrac{1}{2}, 4, -2\right)(21​,4,−2)
  4. (D)(1, 5, 1)

Correct answer: (B)

Step-by-step solution →
Q231·MathematicsSingle correctJEE Main 2019
Let a⃗=i^+j^+2k^,b⃗=b1i^+b2j^+2k^\vec{a} = \hat{i} + \hat{j} + \sqrt{2}\hat{k}, \vec{b} = b_{1}\hat{i} + b_{2}\hat{j} + \sqrt{2}\hat{k}a=i^+j^​+2​k^,b=b1​i^+b2​j^​+2​k^ and c⃗=5i^+j^+2k^\vec{c} = 5\hat{i} + \hat{j} + \sqrt{2}\hat{k}c=5i^+j^​+2​k^ be three vectors such that the projection vector of b⃗\vec{b}b on a⃗\vec{a}a is a⃗\vec{a}a. If a⃗+b⃗\vec{a} + \vec{b}a+b is perpendicular to c⃗\vec{c}c, then ∣b⃗∣\left| \vec{b} \right|​b​ is equal to:
  1. (A)22\sqrt{22}22​
  2. (B)444
  3. (C)32\sqrt{32}32​
  4. (D)666

Correct answer: (D)

Step-by-step solution →
Q232·MathematicsSingle correctJEE Main 2019
Let a⃗=i^−j^\vec a = \hat i - \hat ja=i^−j^​, b⃗=i^+j^+k^\vec b = \hat i + \hat j + \hat kb=i^+j^​+k^ and c⃗\vec cc be a vector such that a⃗×c⃗+b⃗=0\vec a \times \vec c + \vec b = 0a×c+b=0 and a⃗.c⃗=4\vec a . \vec c = 4a.c=4, then ∣c⃗∣2|\vec c|^2∣c∣2 is equal to:
  1. (A)192\dfrac{19}{2}219​
  2. (B)9
  3. (C)8
  4. (D)172\dfrac{17}{2}217​

Correct answer: (A)

Step-by-step solution →
Q233·MathematicsNumericalJEE Advanced 2018
Consider the cube in the first octant with sides OP, OQ and OR of length 1, along the x-axis, y-axis and z-axis, respectively, where O(0, 0, 0) is the origin. Let S(12,12,12)S\left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right)S(21​,21​,21​) be the centre of the cube and T be the vertex of the cube opposite to the origin O such that S lies on the diagonal OT. If p⃗=SP→\vec{p} = \overrightarrow{SP}p​=SP, q⃗=SQ→\vec{q} = \overrightarrow{SQ}q​=SQ​, r⃗=SR→\vec{r} = \overrightarrow{SR}r=SR and t⃗=ST→\vec{t} = \overrightarrow{ST}t=ST, then the value of ∣(p⃗×q⃗)×(r⃗×t⃗)∣\left|(\vec{p} \times \vec{q}) \times (\vec{r} \times \vec{t})\right|​(p​×q​)×(r×t)​ is ______ .

Correct answer: 0.5

Step-by-step solution →
Q234·MathematicsNumericalJEE Advanced 2018
Let a⃗\vec{a}a and b⃗\vec{b}b be two unit vectors such that a⃗⋅b⃗=0\vec{a} \cdot \vec{b} = 0a⋅b=0. For some x, y ∈\in∈ R, let c⃗=xa⃗+yb⃗+(a⃗×b⃗)\vec{c} = x\vec{a} + y\vec{b} + \left(\vec{a} \times \vec{b}\right)c=xa+yb+(a×b). If ∣c⃗∣=2|\vec{c}| = 2∣c∣=2 and the vector c⃗\vec{c}c is inclined at the same angle α\alphaα to both a⃗\vec{a}a and b⃗\vec{b}b, then the value of 8cos⁡2α8\cos^{2}\alpha8cos2α is ______ .

Correct answer: 3

Step-by-step solution →
Q235·MathematicsSingle correctJEE Advanced 2017
Let OOO be the origin and let PQRPQRPQR be an arbitrary triangle. The point SSS is such that OP→⋅OQ→+OR→⋅OS→=OR→⋅OP→+OQ→⋅OS→=OQ→⋅OR→+OP→⋅OS→\overrightarrow{OP}\cdot\overrightarrow{OQ} + \overrightarrow{OR}\cdot\overrightarrow{OS} = \overrightarrow{OR}\cdot\overrightarrow{OP} + \overrightarrow{OQ}\cdot\overrightarrow{OS} = \overrightarrow{OQ}\cdot\overrightarrow{OR} + \overrightarrow{OP}\cdot\overrightarrow{OS}OP⋅OQ​+OR⋅OS=OR⋅OP+OQ​⋅OS=OQ​⋅OR+OP⋅OS Then the triangle PQRPQRPQR has SSS as its
  1. (A)centroid
  2. (B)circumcentre
  3. (C)incentre
  4. (D)orthocenter

Correct answer: (D)

Step-by-step solution →
Q236·MathematicsSingle correctJEE Advanced 2017
PARAGRAPH 1 Let OOO be the origin, and OX→\overrightarrow{OX}OX, OY→\overrightarrow{OY}OY, OZ→\overrightarrow{OZ}OZ be three unit vectors in the directions of the sides QR→\overrightarrow{QR}QR​, RP→\overrightarrow{RP}RP, PQ→\overrightarrow{PQ}PQ​, respectively, of a triangle PQRPQRPQR. ∣OX→×OY→∣=\left|\overrightarrow{OX} \times \overrightarrow{OY}\right| =​OX×OY​=
  1. (A)sin⁡(P+Q)\sin(P + Q)sin(P+Q)
  2. (B)sin⁡2R\sin 2Rsin2R
  3. (C)sin⁡(P+R)\sin(P + R)sin(P+R)
  4. (D)sin⁡(Q+R)\sin(Q + R)sin(Q+R)

Correct answer: (A)

Step-by-step solution →
Q237·MathematicsMultiple correctJEE Advanced 2016
Let u^=u1i^+u2j^+u3k^\hat{u} = u_{1}\hat{i} + u_{2}\hat{j} + u_{3}\hat{k}u^=u1​i^+u2​j^​+u3​k^ be a unit vector in R3\mathbb{R}^{3}R3 and w^=16(i^+j^+2k^)\hat{w} = \frac{1}{\sqrt{6}} \left( \hat{i} + \hat{j} + 2\hat{k} \right)w^=6​1​(i^+j^​+2k^). Given that there exists a vector v⃗\vec{v}v in R3\mathbb{R}^{3}R3 such that ∣u^×v⃗∣=1|\hat{u} \times \vec{v}| = 1∣u^×v∣=1 and w^⋅(u^×v⃗)=1\hat{w} \cdot (\hat{u} \times \vec{v}) = 1w^⋅(u^×v)=1. Which of the following statement(s) is(are) correct?
  1. (A)There is exactly one choice for such v⃗\vec{v}v
  2. (B)There are infinitely many choices for such v⃗\vec{v}v
  3. (C)If u^\hat{u}u^ lies in the xy-plane then ∣u1∣=∣u2∣|u_{1}| = |u_{2}|∣u1​∣=∣u2​∣
  4. (D)If u^\hat{u}u^ lies in the xz-plane then 2∣u1∣=∣u3∣2|u_{1}| = |u_{3}|2∣u1​∣=∣u3​∣

Correct answer: (B), (C)

Step-by-step solution →
Q238·MathematicsMultiple correctJEE Advanced 2015
Let ΔPQR\Delta PQRΔPQR be a triangle. Let a⃗=QR→\vec{a} = \overrightarrow{QR}a=QR​, b⃗=RP→\vec{b} = \overrightarrow{RP}b=RP and c⃗=PQ→\vec{c} = \overrightarrow{PQ}c=PQ​. If ∣a⃗∣=12|\vec{a}| = 12∣a∣=12, ∣b⃗∣=43|\vec{b}| = 4\sqrt{3}∣b∣=43​ and b⃗⋅c⃗=24\vec{b} \cdot \vec{c} = 24b⋅c=24, then which of the following is (are) true ?
  1. (A)∣c⃗∣22−∣a⃗∣=12\dfrac{|\vec{c}|^{2}}{2} - |\vec{a}| = 122∣c∣2​−∣a∣=12
  2. (B)∣c⃗∣22+∣a⃗∣=30\dfrac{|\vec{c}|^{2}}{2} + |\vec{a}| = 302∣c∣2​+∣a∣=30
  3. (C)∣a⃗×b⃗+c⃗×a⃗∣=483|\vec{a} \times \vec{b} + \vec{c} \times \vec{a}| = 48\sqrt{3}∣a×b+c×a∣=483​
  4. (D)a⃗⋅b⃗=−72\vec{a} \cdot \vec{b} = -72a⋅b=−72

Correct answer: (A), (C), (D)

Step-by-step solution →
Q239·MathematicsIntegerJEE Advanced 2015
Suppose that p⃗\vec{p}p​, q⃗\vec{q}q​ and r⃗\vec{r}r are three non-coplanar vectors in R3\mathbb{R}^{3}R3. Let the components of a vector s⃗\vec{s}s along p⃗\vec{p}p​, q⃗\vec{q}q​ and r⃗\vec{r}r be 4, 3 and 5, respectively. If the components of this vector s⃗\vec{s}s along (−p⃗+q⃗+r⃗)\left(-\vec{p} + \vec{q} + \vec{r}\right)(−p​+q​+r), (p⃗−q⃗+r⃗)\left(\vec{p} - \vec{q} + \vec{r}\right)(p​−q​+r) and (−p⃗−q⃗+r⃗)\left(-\vec{p} - \vec{q} + \vec{r}\right)(−p​−q​+r) are xxx, yyy and zzz, respectively, then the value of 2x+y+z2x + y + z2x+y+z is

Correct answer: 9

Step-by-step solution →
Q240·MathematicsMatrix matchJEE Advanced 2015
Match the entries in Column I with the entries in Column II.
Column – IColumn – II
A.In R2\mathbb{R}^{2}R2, if the magnitude of the projection vector of the vector αi^+βj^\alpha\hat{i} + \beta\hat{j}αi^+βj^​ on 3i^+j^\sqrt{3}\hat{i} + \hat{j}3​i^+j^​ is 3\sqrt{3}3​ and if α=2+3β\alpha = 2 + \sqrt{3}\betaα=2+3​β, then possible value(s) of ∣α∣|\alpha|∣α∣ is (are)P.1
B.Let aaa and bbb be real numbers such that the function f(x)={−3ax2−2,x<1bx+a2,x≥1f(x) = \begin{cases} -3ax^{2} - 2, & x < 1 \\ bx + a^{2}, & x \ge 1 \end{cases}f(x)={−3ax2−2,bx+a2,​x<1x≥1​ is differentiable for all x∈Rx \in \mathbb{R}x∈R. Then possible value(s) of aaa is (are)Q.2
C.Let ω≠1\omega \ne 1ω=1 be a complex cube root of unity. If (3−3ω+2ω2)4n+3+(2+3ω−3ω2)4n+3+(−3+2ω+3ω2)4n+3=0(3 - 3\omega + 2\omega^{2})^{4n+3} + (2 + 3\omega - 3\omega^{2})^{4n+3} + (-3 + 2\omega + 3\omega^{2})^{4n+3} = 0(3−3ω+2ω2)4n+3+(2+3ω−3ω2)4n+3+(−3+2ω+3ω2)4n+3=0, then possible value(s) of nnn is (are)R.3
D.Let the harmonic mean of two positive real numbers aaa and bbb be 4. If qqq is a positive real number such that aaa, 5, qqq, bbb is an arithmetic progression, then the value(s) of ∣q−a∣|q - a|∣q−a∣ is (are)S.4
T.5

Correct answer: A-(P,Q); B-(P,Q); C-(P,Q,S,T); D-(Q,T)

Step-by-step solution →
Q241·MathematicsMultiple correctJEE Advanced 2014
Let x⃗\vec{x}x, y⃗\vec{y}y​ and z⃗\vec{z}z be three vectors each of magnitude 2\sqrt{2}2​ and the angle between each pair of them is π3\frac{\pi}{3}3π​. If a⃗\vec{a}a is a non-zero vector perpendicular to x⃗\vec{x}x and y⃗×z⃗\vec{y} \times \vec{z}y​×z and b⃗\vec{b}b is a non-zero vector perpendicular to y⃗\vec{y}y​ and z⃗×x⃗\vec{z} \times \vec{x}z×x, then
  1. (A)b⃗=(b⃗⋅z⃗)(z⃗−x⃗)\vec{b} = (\vec{b} \cdot \vec{z})(\vec{z} - \vec{x})b=(b⋅z)(z−x)
  2. (B)a⃗=(a⃗⋅y⃗)(y⃗−z⃗)\vec{a} = (\vec{a} \cdot \vec{y})(\vec{y} - \vec{z})a=(a⋅y​)(y​−z)
  3. (C)a⃗⋅b⃗=−(a⃗⋅y⃗)(b⃗⋅z⃗)\vec{a} \cdot \vec{b} = -(\vec{a} \cdot \vec{y})(\vec{b} \cdot \vec{z})a⋅b=−(a⋅y​)(b⋅z)
  4. (D)a⃗=(a⃗⋅y⃗)(z⃗−y⃗)\vec{a} = (\vec{a} \cdot \vec{y})(\vec{z} - \vec{y})a=(a⋅y​)(z−y​)

Correct answer: (A), (B), (C)

Step-by-step solution →
Q242·MathematicsIntegerJEE Advanced 2014
Let a⃗\vec{a}a, b⃗\vec{b}b, and c⃗\vec{c}c be three non-coplanar unit vectors such that the angle between every pair of them is π3\frac{\pi}{3}3π​. If a⃗×b⃗+b⃗×c⃗=pa⃗+qb⃗+rc⃗\vec{a} \times \vec{b} + \vec{b} \times \vec{c} = p\vec{a} + q\vec{b} + r\vec{c}a×b+b×c=pa+qb+rc, where ppp, qqq and rrr are scalars, then the value of p2+2q2+r2q2\frac{p^2 + 2q^2 + r^2}{q^2}q2p2+2q2+r2​ is __________

Correct answer: 4

Step-by-step solution →
Q243·MathematicsIntegerJEE Advanced 2013
Consider the set of eight vectors V={ai^+bj^+ck^;a,b,c∈{−1,1}}V=\{a\hat{i}+b\hat{j}+c\hat{k}; a,b,c\in\{-1,1\}\}V={ai^+bj^​+ck^;a,b,c∈{−1,1}}. Three non-coplanar vectors can be chosen from V in 2p2^p2p ways. Then p is ________

Correct answer: 5

Step-by-step solution →
Q244·MathematicsSingle correctJEE Advanced 2013
Match List I with List II and select the correct answer using the code given below the lists :
List-IList-II
P.Volume of parallelepiped determined by vectors a⃗,b⃗\vec{a}, \vec{b}a,b and c⃗\vec{c}c is 2. Then the volume of the parallelepiped determined by vectors 2(a⃗×b⃗),3(b⃗×c⃗)2\left(\vec{a} \times \vec{b}\right), 3\left(\vec{b} \times \vec{c}\right)2(a×b),3(b×c) and (c⃗×a⃗)\left(\vec{c} \times \vec{a}\right)(c×a) is1.100100100
Q.Volume of parallelepiped determined by vectors a⃗,b⃗\vec{a}, \vec{b}a,b and c⃗\vec{c}c is 5. Then the volume of the parallelepiped determined by vectors 3(a⃗+b⃗),(b⃗+c⃗)3\left(\vec{a} + \vec{b}\right), \left(\vec{b} + \vec{c}\right)3(a+b),(b+c) and 2(c⃗+a⃗)2\left(\vec{c} + \vec{a}\right)2(c+a) is2.303030
R.Area of a triangle with adjacent sides determined by vectors a⃗\vec{a}a and b⃗\vec{b}b is 20. Then the area of the triangle with adjacent sides determined by vectors (2a⃗+3b⃗)\left(2\vec{a} + 3\vec{b}\right)(2a+3b) and (a⃗−b⃗)\left(\vec{a} - \vec{b}\right)(a−b) is3.242424
S.Area of a parallelogram with adjacent sides determined by vectors a⃗\vec{a}a and b⃗\vec{b}b is 30. Then the area of the parallelogram with adjacent sides determined by vectors (a⃗+b⃗)\left(\vec{a} + \vec{b}\right)(a+b) and a⃗\vec{a}a is4.606060
  1. (A)P-4, Q-2, R-3, S-1
  2. (B)P-2, Q-3, R-1, S-4
  3. (C)P-3, Q-4, R-1, S-2
  4. (D)P-1, Q-4, R-3, S-2

Correct answer: (C)

Step-by-step solution →
Q245·MathematicsSingle correctJEE Advanced 2013
Let PR→=3i^+j^−2k^\overrightarrow{PR}=3\hat{i}+\hat{j}-2\hat{k}PR=3i^+j^​−2k^ and SQ→=i^−3j^−4k^\overrightarrow{SQ}=\hat{i}-3\hat{j}-4\hat{k}SQ​=i^−3j^​−4k^ determine diagonals of a parallelogram PQRS and PT→=i^+2j^+3k^\overrightarrow{PT}=\hat{i}+2\hat{j}+3\hat{k}PT=i^+2j^​+3k^ be another vector. Then the volume of the parallelepiped determined by the vectors PT→,PQ→\overrightarrow{PT},\overrightarrow{PQ}PT,PQ​ and PS→\overrightarrow{PS}PS is
  1. (A)555
  2. (B)202020
  3. (C)101010
  4. (D)303030

Correct answer: (C)

Step-by-step solution →

Vector Algebra — frequently asked

How many questions from Vector Algebra appear in JEE?

Vector Algebra has appeared in 173 of the last 186 JEE Main and JEE Advanced papers — about 93% of them — contributing 245 questions in total across those papers.

Is Vector Algebra an important chapter for JEE?

Judged by how often it is actually tested, it appears in roughly 93% of papers. Chapters above about 50% are effectively guaranteed to show up every session, so they repay thorough preparation; lower-frequency chapters are better treated as targeted revision.

Where do these Vector Algebra questions come from?

Every question is from an official JEE Main or JEE Advanced paper, transcribed from the original paper and tagged to this chapter. Answers follow the official answer key.

Other Mathematics chapters

  • Three Dimensional Geometry 344
  • Matrices and Determinants 342
  • Sets, Relations and Functions 313
  • Sequence and Series 287
  • Definite Integration 267
  • Differential Equations 240
  • Probability 226
  • Permutations and Combinations 220

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