Vectors PYQs: MHT CET, JEE Main, NEET

Q1MHT CET 2026 · 17 APR · SHIFT 2Easy

Vector Basics, Unit Vectors & Components · Types of vectors & vector laws

There are two vectors $\vec{A} = 6\hat{i} + 9\hat{j} - \hat{k}$ and $\vec{B} = 2\hat{i} + 3\hat{j} - p\hat{k}$ which have the same direction. The value of 'p' is

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Answer: (b)

Vectors in the same direction have proportional components: $\frac{6}{2} = \frac{9}{3} = \frac{-1}{-p}$ .
$\frac{6}{2} = 3 \Rightarrow \frac{1}{p} = 3 \Rightarrow p = \frac{1}{3}$
So the answer is (b) $\frac{1}{3}$ .

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Q2MHT CET 2026 · 15 APR · SHIFT 2Easy

Vector Basics, Unit Vectors & Components · Types of vectors & vector laws

Let $\vec{A}$ and $\vec{B}$ be two non-zero vectors of different magnitude. Which one of the following is the correct equation?

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Answer: (c)

Vector addition is commutative: $\vec{A} + \vec{B} = \vec{B} + \vec{A}$ .
(a) is wrong since $\vec{A} \cdot \vec{B} = +\vec{B} \cdot \vec{A}$ ; (b) is wrong since $\vec{A} \times \vec{B} = -\vec{B} \times \vec{A}$ ; (d) is wrong since $\vec{A} - \vec{B} = -(\vec{B} - \vec{A})$ .
So the answer is (c).

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Q3MHT CET 2026 · 18 APR · SHIFT 2Easy

Vector Basics, Unit Vectors & Components · Unit vector & magnitude

If a unit vector is represented by $\vec{U} = 0.9\hat{i} - 0.2\hat{j} + m\hat{k}$ , then the value of m is

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Answer: (b)

For a unit vector $|\vec{U}|^2 = 1$ .
$0.9^2 + 0.2^2 + m^2 = 1 \Rightarrow 0.81 + 0.04 + m^2 = 1$
$m^2 = 0.15 \Rightarrow m = \sqrt{0.15}$
So the answer is (b) $\sqrt{0.15}$ .

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Q4MHT CET 2026 · 15 APR · SHIFT 1Easy

Vector Basics, Unit Vectors & Components · Unit vector & magnitude

A vector $\vec{A}$ when added to the sum of the vectors $(\hat{i} - 2\hat{j} + 2\hat{k})$ and $(-2\hat{i} + \hat{j} - \hat{k})$ gives a unit vector along Y axis. The magnitude of vector $\vec{A}$ is

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Answer: (b)

Sum of the two vectors: $(\hat{i} - 2\hat{j} + 2\hat{k}) + (-2\hat{i} + \hat{j} - \hat{k}) = -\hat{i} - \hat{j} + \hat{k}$
$\vec{A} + (-\hat{i} - \hat{j} + \hat{k}) = \hat{j} \Rightarrow \vec{A} = \hat{i} + 2\hat{j} - \hat{k}$
$|\vec{A}| = \sqrt{1 + 4 + 1} = \sqrt{6}$
So the answer is (b) $\sqrt{6}$ .

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Q5MHT CET 2026 · 15 MAY · SHIFT 2Easy

Vector Basics, Unit Vectors & Components · Components & direction cosines

The angle made by a vector $\vec{B} = 3\hat{i} + 2\hat{j} + 4\hat{k}$ with Y axis is

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Answer: (c)

Direction cosine with Y axis: $\cos\beta = \frac{B_y}{|\vec{B}|}$ .
$|\vec{B}| = \sqrt{9 + 4 + 16} = \sqrt{29}$
$\cos\beta = \frac{2}{\sqrt{29}} \Rightarrow \beta = \cos^{-1}\left(\frac{2}{\sqrt{29}}\right)$
So the answer is (c).

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Q6MHT CET 2026 · 18 APR · SHIFT 1Easy

Vector Basics, Unit Vectors & Components · Components & direction cosines

The angle made by vector $\vec{A} = 2\hat{i} + 3\hat{j}$ with x-axis and that with y-axis are respectively.

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Answer: (b)

$|\vec{A}| = \sqrt{4 + 9} = \sqrt{13}$
With x-axis: $\cos\alpha = \frac{2}{\sqrt{13}}$ ; with y-axis: $\cos\beta = \frac{3}{\sqrt{13}}$
So the answer is (b) $\cos^{-1}\frac{2}{\sqrt{13}},\ \cos^{-1}\frac{3}{\sqrt{13}}$ .

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Q7JEE Main 2026 · 2 APR · SHIFT 1Medium

Vector Basics, Unit Vectors & Components · Components & direction cosines

The velocity of a particle is given as $\vec{v} = -x\hat{i} + 2y\hat{j} - z\hat{k}$ m/s. The magnitude of acceleration at point (1, 2, 4) is ________ m/s2.

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Answer: (b)

Velocity depends on position, so $a_x = v_x\frac{\partial v_x}{\partial x}$ , and similarly for $y$ and $z$ .
$a_x = (-x)(-1) = x$ , $a_y = (2y)(2) = 4y$ , $a_z = (-z)(-1) = z$
At $(1, 2, 4)$ : $\vec{a} = \hat{i} + 8\hat{j} + 4\hat{k}$
$|\vec{a}| = \sqrt{1 + 64 + 16} = 9\ \text{m/s}^2$
So the answer is (b) 9.

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Q8MHT CET 2025 · 20 APR · SHIFT 2Easy

Vector Basics, Unit Vectors & Components · Unit vector & magnitude

Three vectors are expressed as $\vec{a} = 4\hat{i} - \hat{j}, \vec{b} = -3\hat{i} + 2\hat{j}$ and $\vec{c} = -\hat{k}$ . The unit vector along the direction of sum of these vectors is

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Answer: (a)

$\vec{a} + \vec{b} + \vec{c} = (4 - 3)\hat{i} + (-1 + 2)\hat{j} - \hat{k} = \hat{i} + \hat{j} - \hat{k}$
Magnitude $= \sqrt{1 + 1 + 1} = \sqrt{3}$
Unit vector $= \frac{\hat{i} + \hat{j} - \hat{k}}{\sqrt{3}}$
So the answer is (a).

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Q9MHT CET 2025 · 21 APR · SHIFT 2Medium

Vector Basics, Unit Vectors & Components · Unit vector & magnitude

A unit vector in the direction of resultant vector of $\vec{A} = -2\hat{i} + 3\hat{j} + \hat{k}$ and $\vec{B} = \hat{i} + 2\hat{j} - 4\hat{k}$ is

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Answer: (d)

$\vec{A} + \vec{B} = (-2 + 1)\hat{i} + (3 + 2)\hat{j} + (1 - 4)\hat{k} = -\hat{i} + 5\hat{j} - 3\hat{k}$
Magnitude $= \sqrt{1 + 25 + 9} = \sqrt{35}$
So the answer is (d) $\frac{-\hat{i} + 5\hat{j} - 3\hat{k}}{\sqrt{35}}$ .

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Q10JEE Main 2024 · 30 JAN · SHIFT 2Medium

Vector Basics, Unit Vectors & Components · Unit vector & magnitude

A vector has magnitude same as that of $\vec{A} = 3\hat{i} + 4\hat{j}$ and is parallel to $\vec{B} = 4\hat{i} + 3\hat{j}$ . The $x$ and $y$ components of this vector in first quadrant are $x$ and 3 respectively where $x =$ _________.

Numerical answer type. Work it out, then open the answer.

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Answer: 4

$|\vec{A}| = \sqrt{9 + 16} = 5$ ; unit vector along $\vec{B}$ is $\frac{4\hat{i} + 3\hat{j}}{5}$ .
Required vector $= 5 \times \frac{4\hat{i} + 3\hat{j}}{5} = 4\hat{i} + 3\hat{j}$
So $x = 4$ .

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Q11JEE Main 2024 · 8 APR · SHIFT 1Medium

Vector Basics, Unit Vectors & Components · Components & direction cosines

Three vectors $\overrightarrow{OP}, \overrightarrow{OQ}$ and $\overrightarrow{OR}$ each of magnitude $A$ are acting as shown in figure. The resultant of the three vectors is $A\sqrt{x}$ . The value of $x$ is _________.

Numerical answer type. Work it out, then open the answer.

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Answer: 3

$\vec{OP} = A\hat{i}$ , $\vec{OQ} = A\hat{j}$ , $\vec{OR} = \frac{A}{\sqrt{2}}\hat{i} - \frac{A}{\sqrt{2}}\hat{j}$
Sum $= A\left(1 + \frac{1}{\sqrt{2}}\right)\hat{i} + A\left(1 - \frac{1}{\sqrt{2}}\right)\hat{j}$
$R^2 = A^2\left[\left(1 + \frac{1}{\sqrt{2}}\right)^2 + \left(1 - \frac{1}{\sqrt{2}}\right)^2\right] = A^2(2 + 1) = 3A^2$
$R = A\sqrt{3}$ , so $x = 3$ .

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Q12JEE Main 2023 · 11 APR · SHIFT 2Easy

Vector Basics, Unit Vectors & Components · Unit vector & magnitude

When vector $\vec{A} = 2\hat{i} + 3\hat{j} + 2\hat{k}$ is subtracted from vector $\vec{B}$ , it gives a vector equal to $2\hat{j}$ . Then the magnitude of vector $\vec{B}$ will be:

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Answer: (b)

$\vec{B} - \vec{A} = 2\hat{j} \Rightarrow \vec{B} = \vec{A} + 2\hat{j} = 2\hat{i} + 5\hat{j} + 2\hat{k}$
$|\vec{B}| = \sqrt{4 + 25 + 4} = \sqrt{33}$
So the answer is (b) $\sqrt{33}$ .

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Q13JEE Main 2023 · 15 APR · SHIFT 1Easy

Vector Basics, Unit Vectors & Components · Components & direction cosines

A vector in $x - y$ plane makes an angle of $30^\circ$ with $y$ -axis. The magnitude of $y$ -component of vector is $2\sqrt{3}$ . The magnitude of $x$ -component of the vector will be:

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Answer: (b)

Angle with y-axis is $30^\circ$ : $A_y = A\cos30^\circ$ , $A_x = A\sin30^\circ$ .
$A \cdot \frac{\sqrt{3}}{2} = 2\sqrt{3} \Rightarrow A = 4$
$A_x = 4 \times \frac{1}{2} = 2$
So the answer is (b) 2.

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Q14JEE Main 2021 · 27 AUG · SHIFT 1Medium

Vector Basics, Unit Vectors & Components · Components & direction cosines

The resultant of these forces $\overrightarrow{OP}, \overrightarrow{OQ}, \overrightarrow{OR}, \overrightarrow{OS}$ and $\overrightarrow{OT}$ is approximately .......... N. [Take $\sqrt{3} = 1.7,\ \sqrt{2} = 1.4$ Given $\hat{i}$ and $\hat{j}$ unit vectors along x, y axis]

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Answer: (a)

Resolve each force (angles measured from $+x$ ): P at $60^\circ$ , Q at $30^\circ$ , R at $-45^\circ$ , S at $225^\circ$ , T at $150^\circ$ .
$x$ : $20\cos60^\circ + 10\cos30^\circ + 20\cos45^\circ - 15\cos45^\circ - 15\cos30^\circ = 10 + 8.5 + 14 - 10.5 - 12.75 = 9.25$
$y$ : $20\sin60^\circ + 10\sin30^\circ - 20\sin45^\circ - 15\sin45^\circ + 15\sin30^\circ = 17 + 5 - 14 - 10.5 + 7.5 = 5$
So the answer is (a) $9.25\hat{i} + 5\hat{j}$ .

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Q15JEE Main 2021 · 26 AUG · SHIFT 1Medium

Vector Basics, Unit Vectors & Components · Components & direction cosines

The magnitude of vectors $\overrightarrow{OA} \ \overrightarrow{OB}$ , and $\overrightarrow{OC}$ in the given figure are equal. The direction of $\overrightarrow{OA} + \overrightarrow{OB} - \overrightarrow{OC}$ with x-axis will be:

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Answer: (a)

Take each magnitude as 1. $\vec{OA} = \frac{\sqrt{3}}{2}\hat{i} + \frac{1}{2}\hat{j}$ , $\vec{OB} = \frac{1}{2}\hat{i} - \frac{\sqrt{3}}{2}\hat{j}$ , $\vec{OC} = -\frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j}$
$x$ : $\frac{\sqrt{3}}{2} + \frac{1}{2} + \frac{1}{\sqrt{2}} = \frac{1 + \sqrt{3} + \sqrt{2}}{2}$
$y$ : $\frac{1}{2} - \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} = \frac{1 - \sqrt{3} - \sqrt{2}}{2}$
$\tan\theta = \frac{y}{x} = \frac{1 - \sqrt{3} - \sqrt{2}}{1 + \sqrt{3} + \sqrt{2}}$
So the answer is (a).

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Q16MHT CET 2020 · 19 OCT · SHIFT 2Easy

Vector Basics, Unit Vectors & Components · Components & direction cosines

The angle subtended by the vector $A = 4\hat{i} + 3\hat{j} + 12\hat{k}$ with the $X$ -axis is

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Answer: (d)

$|\vec{A}| = \sqrt{16 + 9 + 144} = 13$
$\cos\alpha = \frac{A_x}{|\vec{A}|} = \frac{4}{13}$
So the answer is (d) $\cos^{-1}\left(\frac{4}{13}\right)$ .

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Q17MHT CET 2020 · 16 OCT · SHIFT 2Easy

Vector Basics, Unit Vectors & Components · Components & direction cosines

The $x, y$ components of vector $\mathbf{P}$ have magnitudes 1 and 3 and $x, y$ components of resultant of $\mathbf{P}$ and $\mathbf{Q}$ have magnitudes 5 and 6, respectively. What is the magnitude of $\mathbf{Q}$ ?

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Answer: (a)

$\vec{Q} = \vec{R} - \vec{P}$ , so subtract components.
$Q_x = 5 - 1 = 4$ , $Q_y = 6 - 3 = 3$
$|\vec{Q}| = \sqrt{16 + 9} = 5$
So the answer is (a) 5.

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Q18MHT CET 2019 · 3 MAY · SHIFT 1Easy

Vector Basics, Unit Vectors & Components · Components & direction cosines

A vector $P$ has $X$ and $Y$ components of magnitude 2 units and 4 units respectively. A vector $Q$ along negative $X$ -axis has magnitude 6 units. The vector $(\mathbf{Q} - \mathbf{P})$ will be

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Answer: (d)

$\vec{P} = 2\hat{i} + 4\hat{j}$ , $\vec{Q} = -6\hat{i}$
$\vec{Q} - \vec{P} = -8\hat{i} - 4\hat{j} = -4(2\hat{i} + \hat{j})$
So the answer is (d).

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Q19JEE Main 2019 · 10 JAN · SHIFT 1Medium

Vector Basics, Unit Vectors & Components · Components & direction cosines

In the cube of side ‘a’ shown in the figure, the vector from the central point of the face ABOD to the central point of the face BEFO will be -

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Answer: (b)

From the figure: $x$ along OD, $y$ along OF, $z$ along OB.
Face ABOD lies in the $xz$ plane: centre $G = \left(\frac{a}{2}, 0, \frac{a}{2}\right)$ .
Face BEFO lies in the $yz$ plane: centre $H = \left(0, \frac{a}{2}, \frac{a}{2}\right)$ .
$\vec{GH} = H - G = \frac{a}{2}(\hat{j} - \hat{i})$
So the answer is (b).

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Q20NEET 1999 · AIPMTEasy

Vector Basics, Unit Vectors & Components · Unit vector & magnitude

If a unit vector is represented by $0.5\hat{i} + 0.8\hat{j} + c\hat{k}$ , the value of $c$ is

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Answer: (b)

For a unit vector the sum of squares of components is 1.
$0.25 + 0.64 + c^2 = 1 \Rightarrow c^2 = 0.11$
So the answer is (b) $\sqrt{0.11}$ .

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Q21MHT CET 2026 · 17 APR · SHIFT 1Easy

Addition & Subtraction of Vectors · Resultant of two vectors

Resultant of two vectors $\vec{P}$ and $\vec{Q}$ is of magnitude A. If $\vec{Q}$ is reversed, then the resultant is of magnitude B. The value of $A^2 + B^2$ is

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Answer: (c)

$A^2 = P^2 + Q^2 + 2PQ\cos\theta$
Reversing $\vec{Q}$ : $B^2 = P^2 + Q^2 - 2PQ\cos\theta$
Adding: $A^2 + B^2 = 2(P^2 + Q^2)$
So the answer is (c).

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Q22MHT CET 2026 · 20 MAY · SHIFT 1Easy

Addition & Subtraction of Vectors · Subtraction of vectors

Let $\vec{P} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{Q} = -\hat{i} - \hat{j} - \hat{k}$ . The angle between $(\vec{P} - \vec{Q})$ and $\vec{P}$ is

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Answer: (d)

$\vec{Q} = -\vec{P}$ , so $\vec{P} - \vec{Q} = 2\vec{P} = 2\hat{i} + 2\hat{j} + 2\hat{k}$ .
$2\vec{P}$ points along $\vec{P}$ , so the angle is $0^\circ$ .
So the answer is (d) $0^\circ$ .

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Q23MHT CET 2026 · 16 APR · SHIFT 1Easy

Addition & Subtraction of Vectors · Subtraction of vectors

Let $\vec{P} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{Q} = -(\hat{i} + \hat{j} + \hat{k})$ . The angle between $(\vec{P} - \vec{Q})$ and $\vec{P}$ is

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Answer: (c)

$\vec{Q} = -\vec{P}$ , so $\vec{P} - \vec{Q} = 2\vec{P}$ .
$2\vec{P}$ is parallel to $\vec{P}$ , so the angle is $0^\circ$ .
So the answer is (c) $0^\circ$ .

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Q24MHT CET 2026 · 14 MAY · SHIFT 2Medium

Addition & Subtraction of Vectors · Resultant of two vectors

The resultant $\vec{R}$ of $\vec{P}$ and $\vec{Q}$ is perpendicular to $\vec{P}$ . Also $|\vec{P}| = |\vec{R}|$ . The angle between $\vec{P}$ and $\vec{Q}$ is ( $\tan 45^\circ = 1$ )

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Answer: (b)

Take $\vec{P} = P\hat{i}$ . $\vec{R} \perp \vec{P}$ and $|\vec{R}| = P$ , so $\vec{R} = P\hat{j}$ .
$\vec{Q} = \vec{R} - \vec{P} = -P\hat{i} + P\hat{j}$
$\vec{Q}$ points at $135^\circ$ to $+x$ , i.e. to $\vec{P}$ .
So the answer is (b) $\frac{3\pi}{4}$ .

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Q25MHT CET 2026 · 14 MAY · SHIFT 1Medium

Addition & Subtraction of Vectors · Resultant of two vectors

A vector $\vec{A}$ has magnitude $6$ units is added to vector $\vec{B}$ which is along $x$ -axis. The resultant of $\vec{A}$ and $\vec{B}$ is along $y$ -axis. If the magnitude of the resultant of $\vec{A}$ and $\vec{B}$ is $3$ times that of $\vec{B}$ then the magnitude of $\vec{B}$ is

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Answer: (b)

$\vec{B} = B\hat{i}$ and $\vec{R} = 3B\hat{j}$ .
$\vec{A} = \vec{R} - \vec{B} = -B\hat{i} + 3B\hat{j}$
$|\vec{A}|^2 = 10B^2 = 36 \Rightarrow B^2 = 3.6$
So the answer is (b) $\sqrt{3.6}$ .

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Q26MHT CET 2026 · 12 MAY · SHIFT 2Medium

Addition & Subtraction of Vectors · Resultant of two vectors

The resultant of two vectors $\vec{A}$ and $\vec{B}$ is $\vec{C}$ . If the magnitude of $\vec{B}$ is doubled, the new resultant vector becomes perpendicular to $\vec{A}$ . Then the magnitude of $\vec{C}$ is

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Answer: (a)

$(\vec{A} + 2\vec{B}) \perp \vec{A}$ : $A^2 + 2AB\cos\theta = 0$ .
$C^2 = A^2 + B^2 + 2AB\cos\theta = (A^2 + 2AB\cos\theta) + B^2 = B^2$
$C = B$
So the answer is (a) $B$ .

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Q27MHT CET 2025 · 19 APR · SHIFT 2Medium

Addition & Subtraction of Vectors · Resultant of two vectors

The resultant of two vectors $\vec{A}$ and $\vec{B}$ is $\vec{C}$ . If the magnitude of $\vec{B}$ is doubled, the new resultant vector becomes perpendicular to $\vec{A}$ , then the magnitude of $\vec{C}$ is

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Answer: (c)

$(\vec{A} + 2\vec{B}) \cdot \vec{A} = 0 \Rightarrow A^2 + 2AB\cos\theta = 0$
$C^2 = A^2 + B^2 + 2AB\cos\theta = B^2 \Rightarrow C = B$
So the answer is (c) B.

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Q28MHT CET 2025 · 21 APR · SHIFT 1Medium

Addition & Subtraction of Vectors · Polygon law & triangle of vectors

The three vector $\vec{A} = 3\hat{i} - 2\hat{j} + \hat{k}, \vec{B} = \hat{i} - 3\hat{j} + 5\hat{k}$ and $\vec{C} = 2\hat{i} - \hat{j} + 4\hat{k}$ will form

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Answer: (c)

Three vectors form a triangle only if one of them equals the sum (or difference) of the other two, i.e. $\vec{A} \pm \vec{B} \pm \vec{C} = 0$ for some signs.
$\vec{B} + \vec{C} = 3\hat{i} - 4\hat{j} + 9\hat{k} \neq \vec{A}$ ; $\vec{A} - \vec{B} = 2\hat{i} + \hat{j} - 4\hat{k} \neq \vec{C}$ ; $\vec{A} + \vec{C} = 5\hat{i} - 3\hat{j} + 5\hat{k} \neq \vec{B}$ .
No combination gives zero, so the vectors do not close into a triangle.
So the answer is (c) no triangle.

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Q29JEE Main 2024 · 5 APR · SHIFT 1Easy

Addition & Subtraction of Vectors · Resultant of two vectors

The angle between vector $\vec{Q}$ and the resultant of $(2\vec{Q} + 2\vec{P})$ and $(2\vec{Q} - 2\vec{P})$ is:

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Answer: (b)

Resultant $= (2\vec{Q} + 2\vec{P}) + (2\vec{Q} - 2\vec{P}) = 4\vec{Q}$
$4\vec{Q}$ is parallel to $\vec{Q}$ , so the angle is zero.
So the answer is (b) $0^\circ$ .

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Q30JEE Main 2024 · 31 JAN · SHIFT 2Easy

Addition & Subtraction of Vectors · Resultant of two vectors

If two vectors $\vec{A}$ and $\vec{B}$ having equal magnitude $R$ are inclined at angle $\theta$ , then

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Answer: (a)

$|\vec{A} + \vec{B}|^2 = 2R^2(1 + \cos\theta) = 4R^2\cos^2\frac{\theta}{2}$
$|\vec{A} + \vec{B}| = 2R\cos\frac{\theta}{2}$ (and $|\vec{A} - \vec{B}| = 2R\sin\frac{\theta}{2}$ , so (b), (c), (d) are wrong)
So the answer is (a).

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Q31JEE Main 2024 · 9 APR · SHIFT 2Medium

Addition & Subtraction of Vectors · Resultant of two vectors

The resultant of two vectors $\vec{A}$ and $\vec{B}$ is perpendicular to $\vec{A}$ and its magnitude is half that of $\vec{B}$ . The angle between vectors $\vec{A}$ and $\vec{B}$ is _________ ${}^\circ$ .

Numerical answer type. Work it out, then open the answer.

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Answer: 150

Resultant $\perp \vec{A}$ : $A + B\cos\theta = 0$ , so $\cos\theta < 0$ (obtuse angle).
Then $R = B\sin\theta$ ; given $R = \frac{B}{2} \Rightarrow \sin\theta = \frac{1}{2}$ .
Obtuse angle with $\sin\theta = \frac{1}{2}$ is $\theta = 150^\circ$ .
Answer: $150^\circ$

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Q32JEE Main 2024 · 9 APR · SHIFT 1Medium

Addition & Subtraction of Vectors · Resultant of two vectors

If $\vec{a}$ and $\vec{b}$ make an angle $\cos^{-1}\left(\frac{5}{9}\right)$ with each other, then $|\vec{a} + \vec{b}| = \sqrt{2}|\vec{a} - \vec{b}|$ for $|\vec{a}| = n|\vec{b}|$ . The integer value of $n$ is _________.

Numerical answer type. Work it out, then open the answer.

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Answer: 3

Square both sides: $a^2 + b^2 + 2ab\cos\theta = 2(a^2 + b^2 - 2ab\cos\theta)$ , so $6ab\cos\theta = a^2 + b^2$ .
Put $\cos\theta = \frac{5}{9}$ and $a = nb$ : $\frac{10}{3}n = n^2 + 1$ .
$3n^2 - 10n + 3 = 0 \Rightarrow n = 3$ or $\frac{1}{3}$ .
Integer value: $n = 3$

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Q33JEE Main 2024 · 4 APR · SHIFT 1Medium

Addition & Subtraction of Vectors · Resultant of two vectors

Two forces $\overline{F}_1$ and $\overline{F}_2$ are acting on a body. One force has magnitude thrice that of the other force and the resultant of the two forces is equal to the force of larger magnitude. The angle between $\vec{F}_1$ and $\vec{F}_2$ is $\cos^{-1}\left(\frac{1}{n}\right)$ . The value of $|n|$ is _______.

Numerical answer type. Work it out, then open the answer.

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Answer: 6

Let the forces be $F$ and $3F$ , resultant $R = 3F$ .
$9F^2 = F^2 + 9F^2 + 6F^2\cos\theta \Rightarrow \cos\theta = -\frac{1}{6}$
So $\theta = \cos^{-1}\left(-\frac{1}{6}\right)$ , i.e. $n = -6$ .
Answer: $|n| = 6$

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Q34JEE Main 2023 · 8 APR · SHIFT 1Easy

Addition & Subtraction of Vectors · Resultant of two vectors

Two forces having magnitude $A$ and $\frac{A}{2}$ are perpendicular to each other. The magnitude of their resultant is:

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Answer: (c)

For perpendicular vectors $R = \sqrt{A^2 + B^2}$ .
$R = \sqrt{A^2 + \frac{A^2}{4}} = \frac{\sqrt{5}A}{2}$
So the answer is (c).

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Q35JEE Main 2022 · 29 JUN · SHIFT 1Medium

Addition & Subtraction of Vectors · Resultant of two vectors

Two vectors $\vec{A}$ and $\vec{B}$ have equal magnitudes. If magnitude of $\vec{A} + \vec{B}$ is equal to two times the magnitude of $\vec{A} - \vec{B}$ , then the angle between $\vec{A}$ and $\vec{B}$ will be:

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Answer: (c)

Equal magnitudes: $|\vec{A} + \vec{B}| = 2A\cos\frac{\theta}{2}$ , $|\vec{A} - \vec{B}| = 2A\sin\frac{\theta}{2}$ .
$\cos\frac{\theta}{2} = 2\sin\frac{\theta}{2} \Rightarrow \tan\frac{\theta}{2} = \frac{1}{2}$
$\cos\theta = \frac{1 - \tan^2(\theta/2)}{1 + \tan^2(\theta/2)} = \frac{1 - 1/4}{1 + 1/4} = \frac{3}{5}$
So the answer is (c) $\cos^{-1}\left(\frac{3}{5}\right)$ .

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Q36JEE Main 2022 · 25 JUN · SHIFT 1Medium

Addition & Subtraction of Vectors · Resultant of two vectors

Which of the following relations is true for two unit vectors $\hat{A}$ and $\hat{B}$ making an angle $\theta$ to each other?

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Answer: (b)

For unit vectors: $|\hat{A} + \hat{B}| = 2\cos\frac{\theta}{2}$ , $|\hat{A} - \hat{B}| = 2\sin\frac{\theta}{2}$ .
Dividing: $\frac{|\hat{A} - \hat{B}|}{|\hat{A} + \hat{B}|} = \tan\frac{\theta}{2}$
So the answer is (b) $|\hat{A} - \hat{B}| = |\hat{A} + \hat{B}|\tan\frac{\theta}{2}$ .

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Q37JEE Main 2021 · 25 JUL · SHIFT 1Easy

Addition & Subtraction of Vectors · Polygon law & triangle of vectors

Match List - I with List - II

Choose the correct answer from the options given below:

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Answer: (b)

Triangle law: when two vectors are drawn head to tail, the vector from the free tail to the free head is their sum.
Rewrite each relation: (a) $\vec{C} = \vec{A} + \vec{B}$ ; (b) $\vec{A} = \vec{B} + \vec{C}$ ; (c) $\vec{B} = \vec{A} + \vec{C}$ ; (d) $\vec{A} + \vec{B} + \vec{C} = 0$ (closed head-to-tail loop).
Match each with the figure in which the single vector on the left runs from the start of the other two to their end; (d) is the figure where all three arrows go round the triangle in one sense.
This gives (a) → (iv), (b) → (iii), (c) → (i), (d) → (ii).
So the answer is (b).

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Q38JEE Main 2021 · 25 JUL · SHIFT 2Medium

Addition & Subtraction of Vectors · Resultant of two vectors

Two vectors $\vec{X}$ and $\vec{Y}$ have equal magnitude. The magnitude of ( $\vec{X} - \vec{Y}$ ) is n times the magnitude of ( $\vec{X} + \vec{Y}$ ). The angle between $\vec{X}$ and $\vec{Y}$ is:

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Answer: (b)

Equal magnitudes: $\frac{|\vec{X} - \vec{Y}|}{|\vec{X} + \vec{Y}|} = \tan\frac{\theta}{2} = n$
$\cos\theta = \frac{1 - n^2}{1 + n^2} = \frac{n^2 - 1}{-n^2 - 1}$
So the answer is (b).

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Q39JEE Main 2021 · 20 JUL · SHIFT 2Medium

Addition & Subtraction of Vectors · Resultant of two vectors

Two vectors $\vec{P}$ and $\vec{Q}$ have equal magnitudes. If the magnitude of $\vec{P} + \vec{Q}$ is n times the magnitude of $\vec{P} - \vec{Q}$ , then angle between $\vec{P}$ and $\vec{Q}$ is:

Show answer and solution

Answer: (d)

Equal magnitudes: $\frac{|\vec{P} + \vec{Q}|}{|\vec{P} - \vec{Q}|} = \cot\frac{\theta}{2} = n$
$\cos\theta = \frac{\cot^2(\theta/2) - 1}{\cot^2(\theta/2) + 1} = \frac{n^2 - 1}{n^2 + 1}$
So the answer is (d) $\cos^{-1}\left(\frac{n^2 - 1}{n^2 + 1}\right)$ .

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Q40JEE Main 2021 · 26 AUG · SHIFT 2Medium

Addition & Subtraction of Vectors · Subtraction of vectors

The angle between vector $\left(\vec{A}\right)$ and $\left(\vec{A} - \vec{B}\right)$ is:

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Answer: (c)

From the figure the angle between $\vec{A}$ and $\vec{B}$ (tail to tail) is $180^\circ - 120^\circ = 60^\circ$ .
Take $\vec{A}$ along $x$ : $\vec{A} - \vec{B} = \left(A - \frac{B}{2}\right)\hat{i} - \frac{\sqrt{3}B}{2}\hat{j}$ .
$\tan\beta = \frac{\sqrt{3}B/2}{A - B/2} = \frac{\sqrt{3}B}{2A - B}$
So the answer is (c) $\tan^{-1}\left(\frac{\sqrt{3}B}{2A - B}\right)$ .

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Q41JEE Main 2021 · 31 AUG · SHIFT 2Difficult

Addition & Subtraction of Vectors · Resultant of two vectors

Statement I:

Two forces $\left(\vec{P} + \vec{Q}\right)$ and $\left(\vec{P} - \vec{Q}\right)$ where $\vec{P} \perp \vec{Q}$ , when act at an angle $\theta$ 1 to each other, the magnitude of their resultant is $\sqrt{3(P^2 + Q^2)}$ , when they act at an angle $\theta$ 2, the magnitude of their resultant becomes $\sqrt{2(P^2 + Q^2)}$ . This is possible only when $\theta_1 < \theta_2$ .

Statement II:

In the situation given above.

$\theta$ 1 = 60 ${}^\circ$ and $\theta$ 2 = 90 ${}^\circ$

In the light of the above statements, choose the most appropriate answer from the options given below :-

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Answer: (b)

Since $\vec{P} \perp \vec{Q}$ : $|\vec{P} + \vec{Q}| = |\vec{P} - \vec{Q}| = \sqrt{P^2 + Q^2} = S$ .
Two equal forces $S$ at angle $\theta$ : $R^2 = 2S^2(1 + \cos\theta)$ .
$R^2 = 3S^2 \Rightarrow \cos\theta_1 = \frac{1}{2},\ \theta_1 = 60^\circ$ ; $R^2 = 2S^2 \Rightarrow \cos\theta_2 = 0,\ \theta_2 = 90^\circ$ .
So $\theta_1 < \theta_2$ and both statements are true: answer (b).

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Q42JEE Main 2021 · 27 JUL · SHIFT 1Difficult

Addition & Subtraction of Vectors · Polygon law & triangle of vectors

Assertion A: If A, B, C, D are four points on a semi-circular arc with centre at 'O' such that , then $\overrightarrow{AB} + \overrightarrow{AC} + \overrightarrow{AD} = 4\overrightarrow{AO} + \overrightarrow{OB} + \overrightarrow{OC}$

Reason R: Polygon law of vector addition yields $\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CD} + \overrightarrow{AD} = 2\overrightarrow{AO}$

In the light of the above statements, choose the most appropriate answer from the options given below:

Show answer and solution

Answer: (a)

Write each vector through O: $\overrightarrow{AX} = \overrightarrow{AO} + \overrightarrow{OX}$ .
$\overrightarrow{AB} + \overrightarrow{AC} + \overrightarrow{AD} = 3\overrightarrow{AO} + \overrightarrow{OB} + \overrightarrow{OC} + \overrightarrow{OD}$ , and $\overrightarrow{OD} = \overrightarrow{AO}$ , giving $4\overrightarrow{AO} + \overrightarrow{OB} + \overrightarrow{OC}$ . A is correct.
Polygon law: $\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CD} = \overrightarrow{AD} = 2\overrightarrow{AO}$ , so adding $\overrightarrow{AD}$ gives $4\overrightarrow{AO}$ , not $2\overrightarrow{AO}$ . R is wrong.
So the answer is (a).

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Q43JEE Main 2021 · 25 FEB · SHIFT 1Difficult

Addition & Subtraction of Vectors · Polygon law & triangle of vectors

In an octagon ABCDEFGH of equal side, what is the sum of

$\overrightarrow{AB} + \overrightarrow{AC} + \overrightarrow{AD} + \overrightarrow{AE} + \overrightarrow{AF} + \overrightarrow{AG} + \overrightarrow{AH}$ ,

if, $\overrightarrow{AO} = 2\hat{i} + 3\hat{j} - 4\hat{k}$

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Answer: (b)

$\overrightarrow{AX} = \overrightarrow{AO} + \overrightarrow{OX}$ for each of the 7 vertices B to H.
Sum $= 7\overrightarrow{AO} + (\overrightarrow{OB} + \dots + \overrightarrow{OH})$ .
For a regular octagon all 8 vectors from the centre add to zero, so $\overrightarrow{OB} + \dots + \overrightarrow{OH} = -\overrightarrow{OA} = \overrightarrow{AO}$ .
Sum $= 8\overrightarrow{AO} = 16\hat{i} + 24\hat{j} - 32\hat{k}$ , answer (b).

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Q44MHT CET 2020 · 16 OCT · SHIFT 1Easy

Addition & Subtraction of Vectors · Resultant of two vectors

Two vectors of same magnitude have a resultant equal to either of the two vectors. The angle between two vectors is

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Answer: (a)

$R^2 = A^2 + A^2 + 2A^2\cos\theta$ with $R = A$ .
$A^2 = 2A^2(1 + \cos\theta) \Rightarrow \cos\theta = -\frac{1}{2}$
So the answer is (a) $\cos^{-1}(-0.5)$ , i.e. $120^\circ$ .

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Q45MHT CET 2020 · 16 OCT · SHIFT 2Medium

Addition & Subtraction of Vectors · Resultant of two vectors

The resultant of two vector $\mathbf{A}$ and $\mathbf{B}$ is $\mathbf{C}$ . If the magnitude of $\mathbf{B}$ is doubled, the new resultant vector becomes perpendicular to A. Then, the magnitude of $\mathbf{C}$ is

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Answer: (b)

$(\vec{A} + 2\vec{B}) \cdot \vec{A} = 0 \Rightarrow A^2 + 2AB\cos\theta = 0$
$C^2 = A^2 + B^2 + 2AB\cos\theta = B^2 \Rightarrow C = B$
So the answer is (b) B.

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Q46JEE Main 2020 · 7 JAN · SHIFT 2Difficult

Addition & Subtraction of Vectors · Resultant of two vectors

The sum of two forces $\vec{P}$ and $\vec{Q}$ is $\vec{R}$ such that $\left|\vec{R}\right| = \left|\vec{P}\right|$ . The angle $\theta$ (in degrees) that the resultant of 2 $\vec{P}$ and $\vec{Q}$ will make with $\vec{Q}$ is ..............

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 90

$|\vec{P} + \vec{Q}|^2 = P^2 \Rightarrow Q^2 + 2\vec{P} \cdot \vec{Q} = 0$ .
$(2\vec{P} + \vec{Q}) \cdot \vec{Q} = 2\vec{P} \cdot \vec{Q} + Q^2 = 0$
Zero dot product means $(2\vec{P} + \vec{Q}) \perp \vec{Q}$ .
Answer: $\theta = 90^\circ$

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Q47MHT CET 2019 · 2 MAY · SHIFT 1Medium

Addition & Subtraction of Vectors · Resultant of two vectors

The resultant $\mathbf{R}$ of $\mathbf{P}$ and $\mathbf{Q}$ is perpendicular to $\mathbf{P}$ . Also $|\mathbf{P}| = |\mathbf{R}|$ . The angle between $\mathbf{P}$ and $\mathbf{Q}$ is $\left[\tan 45^\circ = 1\right]$

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Answer: (d)

Take $\vec{P} = P\hat{i}$ ; then $\vec{R} = P\hat{j}$ (perpendicular, equal magnitude).
$\vec{Q} = \vec{R} - \vec{P} = -P\hat{i} + P\hat{j}$ , which makes $135^\circ$ with $\vec{P}$ .
So the answer is (d) $\frac{3\pi}{4}$ .

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Q48MHT CET 2019 · 2 MAY · SHIFT 1Medium

Addition & Subtraction of Vectors · Resultant of two vectors

If $\sqrt{A^2 + B^2}$ represents the magnitude of resultant of two vectors $(\mathbf{A} + \mathbf{B})$ and $(\mathbf{A} - \mathbf{B})$ , then the angle between two vectors is

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Answer: (c)

Treat the two vectors as having magnitudes $(A + B)$ and $(A - B)$ with angle $\theta$ between them.
$(A + B)^2 + (A - B)^2 + 2(A + B)(A - B)\cos\theta = A^2 + B^2$
$2(A^2 + B^2) + 2(A^2 - B^2)\cos\theta = A^2 + B^2$
$\cos\theta = -\frac{A^2 + B^2}{2(A^2 - B^2)}$
So the answer is (c).

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Q49JEE Main 2019 · 10 JAN · SHIFT 2Medium

Addition & Subtraction of Vectors · Resultant of two vectors

Two vectors $\vec{A}$ and $\vec{B}$ have equal magnitudes. The magnitude of $\left(\vec{A} + \vec{B}\right)$ is 'n' times the magnitude of $\left(\vec{A} - \vec{B}\right)$ . The angle between $\vec{A}$ and $\vec{B}$ is -

Show answer and solution

Answer: (c)

$|\vec{A} + \vec{B}| = 2A\cos\frac{\theta}{2}$ , $|\vec{A} - \vec{B}| = 2A\sin\frac{\theta}{2}$ , so $\cot\frac{\theta}{2} = n$ .
$\cos\theta = \frac{n^2 - 1}{n^2 + 1}$
So the answer is (c) $\cos^{-1}\left[\frac{n^2 - 1}{n^2 + 1}\right]$ .

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Q50JEE Main 2019 · 10 JAN · SHIFT 2Medium

Addition & Subtraction of Vectors · Resultant of two vectors

Two forces P and Q, of magnitude 2F and 3F, respectively, are at an angle $\theta$ with each other. If the force Q is doubled, then their resultant also gets doubled. Then, the angle $\theta$ is -

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Answer: (d)

$R^2 = P^2 + Q^2 + 2PQ\cos\theta$ with $P = 2F$ , $Q = 3F$ : $R^2 = F^2(13 + 12\cos\theta)$ .
With $Q \to 6F$ : $R'^2 = F^2(40 + 24\cos\theta)$ , and $R' = 2R$ means $R'^2 = 4R^2$ .
$40 + 24\cos\theta = 52 + 48\cos\theta \Rightarrow \cos\theta = -\frac{1}{2}$
So the answer is (d) $120^\circ$ .

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Q51JEE Main 2015 · 11 APR · SHIFT 1Medium

Addition & Subtraction of Vectors · Subtraction of vectors

A vector $\vec{A}$ is rotated by a small angle $\Delta\theta$ radian ( $\Delta\theta$ < <1) to get a new vector $\vec{B}$ . In that case is:

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Answer: (c)

$|\vec{B}| = |\vec{A}|$ (rotation does not change length); the angle between them is $\Delta\theta$ .
$|\vec{B} - \vec{A}| = 2|\vec{A}|\sin\frac{\Delta\theta}{2}$
For $\Delta\theta \ll 1$ , $\sin\frac{\Delta\theta}{2} \approx \frac{\Delta\theta}{2}$ , so $|\vec{B} - \vec{A}| \approx |\vec{A}|\Delta\theta$ .
So the answer is (c) $|\vec{A}|\Delta\theta$ .

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Q52MHT CET 2026 · 19 MAY · SHIFT 2Easy

Scalar (Dot) Product · Dot product: value

A vector $\vec{A}$ of magnitude $7\sqrt{2}$ unit, another vector $\vec{B}$ of magnitude $14$ unit are inclined at an angle of $60^\circ$ . The magnitude of the scalar product $(\vec{A} \cdot \vec{B})$ is $\left[\cos(60)^\circ = \frac{1}{2}\right]$

Show answer and solution

Answer: (a)

$\vec{A} \cdot \vec{B} = AB\cos\theta$
$= 7\sqrt{2} \times 14 \times \frac{1}{2} = 49\sqrt{2}$
So the answer is (a) $49\sqrt{2}$ .

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Q53MHT CET 2026 · 19 MAY · SHIFT 1Easy

Scalar (Dot) Product · Dot product: value

In a vector $\vec{p}$ , 'x' and 'y' components have magnitudes 2 units and 5 units respectively. A vector $\vec{q}$ , along negative Y axis has magnitude of 3 units. Then $\vec{p} \cdot \vec{q}$ is

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Answer: (d)

$\vec{p} = 2\hat{i} + 5\hat{j}$ , $\vec{q} = -3\hat{j}$ .
$\vec{p} \cdot \vec{q} = (2)(0) + (5)(-3) = -15$
So the answer is (d) $-15$ .

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Q54MHT CET 2026 · 13 MAY · SHIFT 2Easy

Scalar (Dot) Product · Perpendicular vectors

The vector $\vec{P} = a\hat{i} + a\hat{j} + 3\hat{k}$ and $\vec{Q} = a\hat{i} - 2\hat{j} - \hat{k}$ are perpendicular to each other. Then positive value of ' $a$ ' is

Show answer and solution

Answer: (a)

Perpendicular: $\vec{P} \cdot \vec{Q} = 0$ .
$a^2 - 2a - 3 = 0 \Rightarrow (a - 3)(a + 1) = 0$
Positive value: $a = 3$ , answer (a).

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Q55MHT CET 2026 · 12 MAY · SHIFT 1Easy

Scalar (Dot) Product · Perpendicular vectors

If $\vec{A} = 5\hat{i} + 2\hat{j}$ and $\vec{B} = a\hat{i} + 3\hat{j}$ are perpendicular to each other, the value of 'a' is

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Answer: (a)

$\vec{A} \cdot \vec{B} = 0 \Rightarrow 5a + 6 = 0$
$a = -\frac{6}{5}$ , answer (a).

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Q56MHT CET 2026 · 20 APR · SHIFT 2Easy

Scalar (Dot) Product · Perpendicular vectors

If two vectors $\vec{A} = 4\hat{i} + n\hat{j} + 2\hat{k}$ and $\vec{B} = 2\hat{i} + 2\hat{j} - \hat{k}$ are mutually perpendicular to each other then value of 'n' is

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Answer: (c)

$\vec{A} \cdot \vec{B} = 8 + 2n - 2 = 0$
$2n = -6 \Rightarrow n = -3$ , answer (c).

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Q57MHT CET 2026 · 13 APR · SHIFT 1Easy

Scalar (Dot) Product · Perpendicular vectors

The vector sum of the two forces $\vec{A}$ and $\vec{B}$ is perpendicular to their vector difference. Hence forces $\vec{A}$ and $\vec{B}$ are

Show answer and solution

Answer: (d)

$(\vec{A} + \vec{B}) \cdot (\vec{A} - \vec{B}) = A^2 - B^2 = 0$
So $|\vec{A}| = |\vec{B}|$ : answer (d) equal in magnitude.

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Q58MHT CET 2026 · 11 APR · SHIFT 2Easy

Scalar (Dot) Product · Perpendicular vectors

The vectors $(\vec{A} + \vec{B})$ and $(\vec{A} - \vec{B})$ are perpendicular to each other. This is possible under the condition

Show answer and solution

Answer: (a)

$(\vec{A} + \vec{B}) \cdot (\vec{A} - \vec{B}) = A^2 - B^2 = 0$
So $|\vec{A}| = |\vec{B}|$ , answer (a).

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Q59MHT CET 2026 · 16 APR · SHIFT 2Medium

Scalar (Dot) Product · Perpendicular vectors

Vectors $a\hat{i} + b\hat{j} + \hat{k}$ and $2\hat{i} - 3\hat{j} + 4\hat{k}$ are perpendicular to each other when $3a + 2b = 7$ , the ratio of a to b is $x/2$ . The value of x is

Show answer and solution

Answer: (b)

Perpendicular: $2a - 3b + 4 = 0$ ; also $3a + 2b = 7$ .
Solving: $a = 1$ , $b = 2$ .
$\frac{a}{b} = \frac{1}{2} = \frac{x}{2} \Rightarrow x = 1$
So the answer is (b) 1.

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Q60MHT CET 2026 · 13 APR · SHIFT 2Medium

Scalar (Dot) Product · Perpendicular vectors

Calculate the values of $a$ and $b$ if vectors $a\hat{i} + b\hat{j} = \hat{n}$ and $(\hat{i} + \hat{j})$ are perpendicular to each other

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Answer: (d)

$\hat{n}$ is a unit vector: $a^2 + b^2 = 1$ .
Perpendicular to $\hat{i} + \hat{j}$ : $a + b = 0 \Rightarrow b = -a$ .
$2a^2 = 1 \Rightarrow a = \frac{1}{\sqrt{2}},\ b = -\frac{1}{\sqrt{2}}$
So the answer is (d).

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Q61MHT CET 2025 · 22 APR · SHIFT 2Easy

Scalar (Dot) Product · Dot product: value

Given $\vec{A} = (2\hat{i} - 3\hat{j} + \hat{k}), \quad \vec{B} = (3\hat{i} + \hat{j} - 2\hat{k})$ and $\vec{C} = (3\hat{i} + 2\hat{j} + \hat{k})$ , then $(\vec{A} + \vec{B}) \cdot \vec{C}$ will be

Show answer and solution

Answer: (a)

$\vec{A} + \vec{B} = 5\hat{i} - 2\hat{j} - \hat{k}$
$(\vec{A} + \vec{B}) \cdot \vec{C} = 5(3) + (-2)(2) + (-1)(1) = 15 - 4 - 1 = 10$
So the answer is (a) 10.

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Q62MHT CET 2025 · 22 APR · SHIFT 2Easy

Scalar (Dot) Product · Dot product: value

Given $\vec{A} = (2\hat{i} - 3\hat{j} + \hat{k}), \quad \vec{B} = (3\hat{i} + \hat{j} - 2\hat{k})$ and $\vec{C} = (3\hat{i} + 2\hat{j} + \hat{k})$ , then $(\vec{A} + \vec{B}) \cdot \vec{C}$ will be

Show answer and solution

Answer: (a)

$\vec{A} + \vec{B} = 5\hat{i} - 2\hat{j} - \hat{k}$
$(\vec{A} + \vec{B}) \cdot \vec{C} = 15 - 4 - 1 = 10$
So the answer is (a) 10.

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Q63MHT CET 2025 · 26 APR · SHIFT 2Easy

Scalar (Dot) Product · Perpendicular vectors

The vector sum of two forces $\vec{A}$ and $\vec{B}$ is perpendicular to their vector difference. Hence forces $\vec{A}$ and $\vec{B}$ are

Show answer and solution

Answer: (d)

$(\vec{A} + \vec{B}) \cdot (\vec{A} - \vec{B}) = A^2 - B^2 = 0$
So $|\vec{A}| = |\vec{B}|$ : answer (d) equal in magnitude.

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Q64MHT CET 2025 · 5 MAY · SHIFT 2Medium

Scalar (Dot) Product · Perpendicular vectors

Two vectors $a\hat{i} + b\hat{j} + \hat{k}$ and $2\hat{i} - 3\hat{j} + 4\hat{k}$ are perpendicular to each other. When $3a + 2b = 7$ , the ratio of $a$ to $b$ is $\frac{x}{2}$ . The value of $x$ is

Show answer and solution

Answer: (c)

Perpendicular: $2a - 3b + 4 = 0$ ; also $3a + 2b = 7$ .
Solving: $a = 1$ , $b = 2$ .
$\frac{a}{b} = \frac{1}{2} = \frac{x}{2} \Rightarrow x = 1$
So the answer is (c) 1.

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Q65MHT CET 2025 · 23 APR · SHIFT 1Medium

Scalar (Dot) Product · Perpendicular vectors

If $\vec{P} = b\hat{i} + 6\hat{j} + \hat{k}$ and $\vec{Q} = \hat{i} - a\hat{j} + 4\hat{k}$ are perpendicular to each other, also $3b - a = 5$ . The value of $a$ and $b$ is

Show answer and solution

Answer: (b)

Perpendicular: $b - 6a + 4 = 0 \Rightarrow b = 6a - 4$ .
Put in $3b - a = 5$ : $18a - 12 - a = 5 \Rightarrow a = 1$ .
$b = 6 - 4 = 2$
So the answer is (b) $a = 1,\ b = 2$ .

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Q66JEE Main 2025 · 23 JAN · SHIFT 1Medium

Scalar (Dot) Product · Perpendicular vectors

Two particles are located at equal distance from origin. The position vectors of those are represented by $\vec{A} = 2\hat{i} + 3n\hat{j} + 2\hat{k}$ and $\vec{B} = 2\hat{i} - 2\hat{j} + 4p\hat{k}$ , respectively. If both the vectors are at right angle to each other, the value of $n^{-1}$ is ________.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 3

Equal distance: $4 + 9n^2 + 4 = 4 + 4 + 16p^2 \Rightarrow 9n^2 = 16p^2,\ p = \pm\frac{3n}{4}$ .
Perpendicular: $\vec{A} \cdot \vec{B} = 4 - 6n + 8p = 0$ .
$p = +\frac{3n}{4}$ gives $4 = 0$ (impossible); $p = -\frac{3n}{4}$ gives $4 - 12n = 0 \Rightarrow n = \frac{1}{3}$ .
Answer: $n^{-1} = 3$

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Q67JEE Main 2023 · 24 JAN · SHIFT 2Easy

Scalar (Dot) Product · Perpendicular vectors

If two vectors $\vec{P} = \hat{i} + 2m\hat{j} + m\hat{k}$ and $\vec{Q} = 4\hat{i} - 2\hat{j} + m\hat{k}$ are perpendicular to each other. Then, the value of m will be:

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Answer: (d)

$\vec{P} \cdot \vec{Q} = 4 - 4m + m^2 = 0$
$(m - 2)^2 = 0 \Rightarrow m = 2$ , answer (d).

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Q68JEE Main 2023 · 24 JAN · SHIFT 1Medium

Scalar (Dot) Product · Perpendicular vectors

Vectors $a\hat{i} + b\hat{j} + \hat{k}$ and $2\hat{i} - 3\hat{j} + 4\hat{k}$ are perpendicular to each other when $3a + 2b = 7$ , the ratio of $a$ to $b$ is $\frac{x}{2}$ . The value of $x$ is ____________.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 1

Perpendicular: $2a - 3b + 4 = 0$ ; also $3a + 2b = 7$ .
Solving: $a = 1$ , $b = 2$ .
$\frac{a}{b} = \frac{1}{2} = \frac{x}{2}$
Answer: $x = 1$

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Q69JEE Main 2022 · 28 JUL · SHIFT 1Easy

Scalar (Dot) Product · Angle & projection

If the projection of $2\hat{i} + 4\hat{j} - 2\hat{k}$ on $\hat{i} + 2\hat{j} + \alpha\hat{k}$ is zero. Then, the value of $\alpha$ will be ___________.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 5

Projection zero means the dot product is zero.
$2(1) + 4(2) - 2\alpha = 0 \Rightarrow 10 = 2\alpha$
Answer: $\alpha = 5$

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Q70JEE Main 2022 · 26 JUL · SHIFT 2Easy

Scalar (Dot) Product · Angle & projection

If $\vec{A} = (2\hat{i} + 3\hat{j} - \hat{k})\text{ m}$ and $\vec{B} = (\hat{i} + 2\hat{j} + 2\hat{k})\text{ m}$ . The magnitude of component of vector $\vec{A}$ along vector $\vec{B}$ will be ____________ $\text{m}$ .

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 2

Component of $\vec{A}$ along $\vec{B}$ $= \frac{\vec{A} \cdot \vec{B}}{|\vec{B}|}$ .
$\vec{A} \cdot \vec{B} = 2 + 6 - 2 = 6$ , $|\vec{B}| = \sqrt{1 + 4 + 4} = 3$
Component $= \frac{6}{3} = 2\text{ m}$

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Q71JEE Main 2021 · 22 JUL · SHIFT 2Easy

Scalar (Dot) Product · Angle & projection

What will be the projection of vector $\vec{A} = \hat{i} + \hat{j} + \hat{k}$ on vector $\vec{B} = \hat{i} + \hat{j}$ ?

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Answer: (b)

Vector projection of $\vec{A}$ on $\vec{B}$ : $\left(\frac{\vec{A} \cdot \vec{B}}{B^2}\right)\vec{B}$ .
$\vec{A} \cdot \vec{B} = 2$ , $B^2 = 2$ .
Projection $= \hat{i} + \hat{j}$ , answer (b).

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Q72MHT CET 2019 · 3 MAY · SHIFT 1Easy

Scalar (Dot) Product · Perpendicular vectors

The vectors $(\mathbf{A} + \mathbf{B})$ and $(\mathbf{A} - \mathbf{B})$ are at right angles to each other. This is possible under the condition

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Answer: (a)

$(\mathbf{A} + \mathbf{B}) \cdot (\mathbf{A} - \mathbf{B}) = A^2 - B^2 = 0$
So $|A| = |B|$ , answer (a).

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Q73MHT CET 2019 · 2 MAY · SHIFT 2Easy

Scalar (Dot) Product · Angle & projection

$\mathbf{P}$ and $\mathbf{Q}$ are two non-zero vectors inclined to each other at an angle ' $\theta$ '. ' $p$ ' and ' $q$ ' are unit vectors along $\mathbf{P}$ and $\mathbf{Q}$ respectively. The component of $\mathbf{Q}$ in the direction of $\mathbf{P}$ will be

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Answer: (a)

Component of $\mathbf{Q}$ along $\mathbf{P}$ $= Q\cos\theta$ .
$p \cdot Q = (1)(Q)\cos\theta = Q\cos\theta$ since $p$ is a unit vector along $\mathbf{P}$ .
So the answer is (a) $p \cdot Q$ .

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Q74JEE Main 2019 · 8 APR · SHIFT 2Difficult

Scalar (Dot) Product · Dot product: value

Let $\left|\vec{A}_1\right| = 3, \ \left|\vec{A}_2\right| = 5$ and $\left|\vec{A}_1 + \vec{A}_2\right| = 5$ . The value of $\left(2\vec{A}_1 + 3\vec{A}_2\right) \cdot \left(3\vec{A}_1 - 2\vec{A}_2\right)$ is :-

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Answer: (a)

$|\vec{A}_1 + \vec{A}_2|^2 = 9 + 25 + 2\vec{A}_1 \cdot \vec{A}_2 = 25 \Rightarrow \vec{A}_1 \cdot \vec{A}_2 = -4.5$
$(2\vec{A}_1 + 3\vec{A}_2) \cdot (3\vec{A}_1 - 2\vec{A}_2) = 6A_1^2 + 5\vec{A}_1 \cdot \vec{A}_2 - 6A_2^2$
$= 54 - 22.5 - 150 = -118.5$
So the answer is (a) $-118.5$ .

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Q75JEE Main 2018 · 16 APR · SHIFT 1Difficult

Scalar (Dot) Product · Perpendicular vectors

Let $\vec{A} = (\hat{i} + \hat{j})$ and, $\vec{B} = (2\hat{i} - \hat{j}).$ The magnitude of a coplanar vector $\vec{C}$ such that $\vec{A} \cdot \vec{C} = \vec{B} \cdot \vec{C} = \vec{A} \cdot \vec{B},$ is given by:

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Answer: (b)

$\vec{A} \cdot \vec{B} = 2 - 1 = 1$ . Let $\vec{C} = x\hat{i} + y\hat{j}$ .
$\vec{A} \cdot \vec{C} = x + y = 1$ and $\vec{B} \cdot \vec{C} = 2x - y = 1$ .
$x = \frac{2}{3},\ y = \frac{1}{3} \Rightarrow |\vec{C}| = \sqrt{\frac{4}{9} + \frac{1}{9}} = \sqrt{\frac{5}{9}}$
So the answer is (b).

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Q76MHT CET 2026 · 20 APR · SHIFT 1Medium

Vector (Cross) Product · Cross product & area

The area of parallelogram formed by vectors $\vec{P} = 2\hat{i} - \hat{j} + 5\hat{k}$ and $\vec{Q} = 3\hat{i} - 2\hat{j} + 4\hat{k}$ is

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Answer: (b)

Area of parallelogram $= |\vec{P} \times \vec{Q}|$ .
$\vec{P} \times \vec{Q} = \hat{i}[(-1)(4) - (5)(-2)] - \hat{j}[(2)(4) - (5)(3)] + \hat{k}[(2)(-2) - (-1)(3)]$
$= 6\hat{i} + 7\hat{j} - \hat{k}$
$|\vec{P} \times \vec{Q}| = \sqrt{36 + 49 + 1} = \sqrt{86}$
So the answer is (b) $\sqrt{86}$ .

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Q77MHT CET 2026 · 15 MAY · SHIFT 1Medium

Vector (Cross) Product · Dot–cross relations

If $|\vec{A} \times \vec{B}| = \sqrt{3}(\vec{A} \cdot \vec{B})$ then the value of $|\vec{A} + \vec{B}|$ is ( $\tan 60^\circ = \sqrt{3} \quad \cos 60^\circ = \frac{1}{2}$ )

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Answer: (d)

$AB\sin\theta = \sqrt{3}\,AB\cos\theta \Rightarrow \tan\theta = \sqrt{3} \Rightarrow \theta = 60^\circ$ .
$|\vec{A} + \vec{B}|^2 = A^2 + B^2 + 2AB\cos 60^\circ = A^2 + B^2 + AB$
So the answer is (d) $(A^2 + B^2 + AB)^{1/2}$ .

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Q78MHT CET 2026 · 19 APR · SHIFT 1Medium

Vector (Cross) Product · Dot–cross relations

For two vectors $\vec{P}$ and $\vec{Q} \quad \vec{P} \cdot \vec{Q} = |\vec{P} \times \vec{Q}|$

The magnitude of $\vec{R} = \vec{P} + \vec{Q}$ is ( $\cos 45^\circ = \frac{1}{\sqrt{2}}$ )?

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Answer: (d)

$PQ\cos\theta = PQ\sin\theta \Rightarrow \tan\theta = 1 \Rightarrow \theta = 45^\circ$ .
$R^2 = P^2 + Q^2 + 2PQ\cos 45^\circ = P^2 + Q^2 + \sqrt{2}\,PQ$
So the answer is (d) $\sqrt{P^2 + Q^2 + \sqrt{2}\,PQ}$ .

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Q79MHT CET 2025 · 26 APR · SHIFT 1Easy

Vector (Cross) Product · Cross product & area

If $|\vec{a}| = \sqrt{26}, \ |\vec{b}| = 7$

$|\vec{a} \times \vec{b}| = 35$ , find $\vec{a} \cdot \vec{b}$

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Answer: (d)

$|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = a^2b^2$
$(\vec{a} \cdot \vec{b})^2 = 26 \times 49 - 35^2 = 1274 - 1225 = 49$
$\vec{a} \cdot \vec{b} = 7$ , answer (d).

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Q80MHT CET 2025 · 25 APR · SHIFT 1Easy

Vector (Cross) Product · Cross product & area

Vector $\vec{A}$ of magnitude $5\sqrt{3}$ units, another vector $\vec{B}$ of magnitude of 10 units are inclined to each other at an angle of $30^\circ$ . The magnitude of vector product of the two vectors is $\left[\sin 30^\circ = \frac{1}{2}\right]$

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Answer: (c)

$|\vec{A} \times \vec{B}| = AB\sin\theta$
$= 5\sqrt{3} \times 10 \times \frac{1}{2} = 25\sqrt{3}$ units
So the answer is (c).

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Q81MHT CET 2025 · 20 APR · SHIFT 1Medium

Vector (Cross) Product · Direction & triple product

If $\vec{A} = \hat{i} + \hat{j} + 3\hat{k}, \ \vec{B} = -\hat{i} + \hat{j} + 4\hat{k}$ and $\vec{C} = 2\hat{i} - 2\hat{j} - 8\hat{k}$ , then the angle between the vectors $\vec{P} = \vec{A} + \vec{B} + \vec{C}$ and $\vec{Q} = (\vec{A} \times \vec{B})$ is (in degree)

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Answer: (c)

$\vec{P} = \vec{A} + \vec{B} + \vec{C} = 2\hat{i} + 0\hat{j} - \hat{k}$ .
$\vec{Q} = \vec{A} \times \vec{B} = \hat{i}(4 - 3) - \hat{j}(4 + 3) + \hat{k}(1 + 1) = \hat{i} - 7\hat{j} + 2\hat{k}$ .
$\vec{P} \cdot \vec{Q} = 2 + 0 - 2 = 0$ , so $\vec{P} \perp \vec{Q}$ .
So the answer is (c) $90^\circ$ .

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Q82JEE Main 2024 · 6 APR · SHIFT 1Medium

Vector (Cross) Product · Direction & triple product

For three vectors $\vec{A} = (-x\hat{i} - 6\hat{j} - 2\hat{k}), \ \vec{B} = (-\hat{i} + 4\hat{j} + 3\hat{k})$ and $\vec{C} = (-8\hat{i} - \hat{j} + 3\hat{k})$ , if $\vec{A} \cdot (\vec{B} \times \vec{C}) = 0$ , then value of $x$ is ________.

Numerical answer type. Work it out, then open the answer.

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Answer: 4

$\vec{B} \times \vec{C} = \hat{i}[(4)(3) - (3)(-1)] - \hat{j}[(-1)(3) - (3)(-8)] + \hat{k}[(-1)(-1) - (4)(-8)]$
$= 15\hat{i} - 21\hat{j} + 33\hat{k}$
$\vec{A} \cdot (\vec{B} \times \vec{C}) = -15x + 126 - 66 = 60 - 15x = 0$
So $x = 4$ .

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Q83JEE Main 2023 · 25 JAN · SHIFT 1Medium

Vector (Cross) Product · Direction & triple product

If $\vec{P} = 3\hat{i} + \sqrt{3}\hat{j} + 2\hat{k}$ and $\vec{Q} = 4\hat{i} + \sqrt{3}\hat{j} + 2.5\hat{k}$ then, the unit vector in the direction of $\vec{P} \times \vec{Q}$ is $\frac{1}{x}\left(\sqrt{3}\hat{i} + \hat{j} - 2\sqrt{3}\hat{k}\right)$ . The value of $x$ is _________.

Numerical answer type. Work it out, then open the answer.

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Answer: 4

$\vec{P} \times \vec{Q} = \hat{i}(2.5\sqrt{3} - 2\sqrt{3}) - \hat{j}(7.5 - 8) + \hat{k}(3\sqrt{3} - 4\sqrt{3})$
$= \frac{1}{2}\left(\sqrt{3}\hat{i} + \hat{j} - 2\sqrt{3}\hat{k}\right)$
Magnitude of $\sqrt{3}\hat{i} + \hat{j} - 2\sqrt{3}\hat{k}$ is $\sqrt{3 + 1 + 12} = 4$ .
Unit vector $= \frac{1}{4}\left(\sqrt{3}\hat{i} + \hat{j} - 2\sqrt{3}\hat{k}\right)$ , so $x = 4$ .

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Q84NEET 2022 · PHASE 2Easy

Vector (Cross) Product · Cross product & area

If $\vec{F} = 2\hat{i} + \hat{j} - \hat{k}$ and $\vec{r} = 3\hat{i} + 2\hat{j} - 2\hat{k}$ , then the scalar and vector products of $\vec{F}$ and $\vec{r}$ have the magnitudes respectively as

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Answer: (d)

$\vec{F} \cdot \vec{r} = 6 + 2 + 2 = 10$
$\vec{F} \times \vec{r} = \hat{i}(-2 + 2) - \hat{j}(-4 + 3) + \hat{k}(4 - 3) = \hat{j} + \hat{k}$
$|\vec{F} \times \vec{r}| = \sqrt{2}$
So the answer is (d) $10, \sqrt{2}$ .

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Q85JEE Main 2022 · 25 JUN · SHIFT 1Easy

Vector (Cross) Product · Cross product & area

$\vec{A}$ is a vector quantity such that $|\vec{A}|$ = non-zero constant. Which of the following expression is true for $\vec{A}$ ?

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Answer: (c)

$|\vec{A} \times \vec{A}| = A^2\sin 0^\circ = 0$ , so $\vec{A} \times \vec{A}$ is the null vector.
$\vec{A} \cdot \vec{A} = A^2 \neq 0$ , and a vector cannot be $< 0$ or $> 0$ .
So the answer is (c) $\vec{A} \times \vec{A} = 0$ .

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Q86JEE Main 2021 · 25 FEB · SHIFT 2Easy

Vector (Cross) Product · Cross product & area

If $\vec{P} \times \vec{Q} = \vec{Q} \times \vec{P}$ , the angle between $\vec{P}$ and $\vec{Q}$ is $\theta(0^\circ$ < $\theta$ < 360 $^\circ$ ). The value of ' $\theta$ ' will be ___________ $^\circ$ .

Numerical answer type. Work it out, then open the answer.

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Answer: 180

$\vec{Q} \times \vec{P} = -\vec{P} \times \vec{Q}$ , so the condition gives $2\vec{P} \times \vec{Q} = 0$ .
$PQ\sin\theta = 0 \Rightarrow \sin\theta = 0$ .
In $0^\circ < \theta < 360^\circ$ the only value is $\theta = 180^\circ$ .
So $\theta = 180^\circ$ .

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Q87JEE Main 2021 · 20 JUL · SHIFT 1Medium

Vector (Cross) Product · Dot–cross relations

If $\vec{A}$ and $\vec{B}$ are two vectors satisfying the relation $\vec{A} \quad \vec{B}$ . = . Then the value of will be:

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Answer: (c)

Given $\vec{A} \cdot \vec{B} = |\vec{A} \times \vec{B}|$ , find $|\vec{A} - \vec{B}|$ .
$AB\cos\theta = AB\sin\theta \Rightarrow \theta = 45^\circ$ .
$|\vec{A} - \vec{B}|^2 = A^2 + B^2 - 2AB\cos 45^\circ = A^2 + B^2 - \sqrt{2}AB$
So the answer is (c) $\sqrt{A^2 + B^2 - \sqrt{2}AB}$ .

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Q88JEE Main 2021 · 22 JUL · SHIFT 2Difficult

Vector (Cross) Product · Direction & triple product

Three particles P, Q and R are moving along the vectors $\vec{A} = \hat{i} + \hat{j} \quad \vec{B} = \hat{j} + \hat{k}$ and $\vec{C} = -\hat{i} + \hat{j}$ respectively. They strike on a point and start to move in different directions. Now particle P is moving normal to the plane which contains vector $\vec{A}$ and $\vec{B}$ . Similarly particle Q is moving normal to the plane which contains vector $\vec{A}$ and $\vec{C}$ . The angle between the direction of motion of P and Q is $\cos^{-1}\left(\frac{1}{\sqrt{x}}\right)$ . Then the value of x is _______________.

Numerical answer type. Work it out, then open the answer.

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Answer: 3

Direction of P: $\vec{A} \times \vec{B} = (\hat{i} + \hat{j}) \times (\hat{j} + \hat{k}) = \hat{i} - \hat{j} + \hat{k}$ .
Direction of Q: $\vec{A} \times \vec{C} = (\hat{i} + \hat{j}) \times (-\hat{i} + \hat{j}) = 2\hat{k}$ .
$\cos\theta = \frac{(\hat{i} - \hat{j} + \hat{k}) \cdot 2\hat{k}}{\sqrt{3} \times 2} = \frac{1}{\sqrt{3}}$
So $x = 3$ .

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Q89MHT CET 2020 · 19 OCT · SHIFT 2Easy

Vector (Cross) Product · Cross product & area

What is the angle between resultant of $A + B$ and $\mathbf{A} \times \mathbf{B}$ .

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Answer: (d)

$\mathbf{A} \times \mathbf{B}$ is perpendicular to the plane containing $\mathbf{A}$ and $\mathbf{B}$ .
$\mathbf{A} + \mathbf{B}$ lies in that plane, so $(\mathbf{A} + \mathbf{B}) \cdot (\mathbf{A} \times \mathbf{B}) = 0$ .
So the angle is (d) $\frac{\pi}{2}$ rad.

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Q90JEE Main 2004 · AIEEEEasy

Vector (Cross) Product · Cross product & area

If $\vec{A} \times \vec{B} = \vec{B} \times \vec{A}$ , then the angle between A and B is

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Answer: (c)

Cross product is anti-commutative: $\vec{B} \times \vec{A} = -\vec{A} \times \vec{B}$ .
Given $\vec{A} \times \vec{B} = -\vec{A} \times \vec{B} \Rightarrow 2\vec{A} \times \vec{B} = 0 \Rightarrow AB\sin\theta = 0$ .
$\sin\theta = 0 \Rightarrow \theta = 0$ or $\pi$ ; of the options only $\pi$ fits.
So the answer is (c) $\pi$ .

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Q91NEET 1999 · AIPMTMedium

Vector (Cross) Product · Direction & triple product

What is the linear velocity if angular velocity vector $\vec{\omega} = 3\hat{i} - 4\hat{j} + \hat{k}$ and position vector $\vec{r} = 5\hat{i} - 6\hat{j} + 6\hat{k}$ ?

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Answer: (b)

Linear velocity $\vec{v} = \vec{\omega} \times \vec{r}$ .
$\vec{v} = \hat{i}[(-4)(6) - (1)(-6)] - \hat{j}[(3)(6) - (1)(5)] + \hat{k}[(3)(-6) - (-4)(5)]$
$= -18\hat{i} - 13\hat{j} + 2\hat{k}$
So the answer is (b) $-18\hat{i} - 13\hat{j} + 2\hat{k}$ .

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