Question 1: Palette 1/53

DPT MCQ

Concept: Vector Addition and Resultant Magnitude

Two forces $\vec{A}$ and $\vec{B}$ of magnitude 10 N and 20 N are acting on a block having 60° angle between them. Find the magnitude of resultant of them.

A

$10\sqrt{3}$ N

B

$10\sqrt{5}$ N

C

$10\sqrt{7}$ N

D

30 N

Question 2: Palette 2/53

DPT MCQ

Concept: Vector Addition and Resultant of Orthogonal Forces

Max and min magnitude of resultant of two forces are 7 N and 1 N respectively. Find the resultant of 2 forces when it act orthogonally.

A

3 N

B

4 N

C

5 N

D

7 N

Question 3: Palette 3/53

DPT MCQ

Concept: Vector Addition

Two vectors A & B have the same magnitude x. Find the magnitude of the resultant of A and B if the angle between them is 60°.

A

x

B

$x\sqrt{3}$

C

$x\sqrt{2}$

D

2x

Question 4: Palette 4/53

DPT MCQ

Concept: Vector Addition and Resultant Magnitude

If two vectors A and B of magnitude 10 N and 6 N are at an angle 60°. If B becomes twice its initial value and is added to A. Find the magnitude of the resultant of A and B after the change in B.

A

$\sqrt{244}$ N

B

$\sqrt{364}$ N

C

20 N

D

26 N

Question 5: Palette 5/53

DPT MCQ

Concept: Range of Resultant Vector Magnitude

Magnitude of vector A is 8 N. Magnitude of vector B is 6 N. Which of the following can be the magnitude of A + B?

A

10 N

B

22 N

C

48 N

D

1.99 N

Question 6: Palette 6/53

DPT MCQ

Concept: Vector Addition and Resultant Magnitude

Magnitude of resultant of vectors A and B is 5 unit where magnitude of vector A is $5\sqrt{3}$ unit and magnitude of vector B is 5 unit. Find the angle between vectors A and B.

A

30°

B

60°

C

120°

D

150°

Question 7: Palette 7/53

DPT MCQ

Concept: Vector Addition and Direction of Resultant

Sum of the magnitude of vector A and vector B is 16 N. Magnitude of resultant of vector A and B is 8 N. When resultant is perpendicular to vector A. Find magnitude of A and B.

A

A = 6 N, B = 10 N

B

A = 10 N, B = 6 N

C

A = 8 N, B = 8 N

D

A = 4 N, B = 12 N

Question 8: Palette 8/53

DPT MCQ

Concept: Vector Addition and Direction of Resultant

Resultant of vector A and vector B is perpendicular to vector A and its magnitude is equal to half of the magnitude of vector B. Find the angle between vector A and vector B.

A

30°

B

60°

C

120°

D

150°

Question 9: Palette 9/53

DPT MCQ

Concept: Vector Addition and Resultant Magnitude

Two vectors A and B have the same magnitude 'a' and their resultant has magnitude R. Now B is doubled and added to A and the new resultant becomes $a\sqrt{3}$. Find the angle between A and B.

A

60°

B

90°

C

120°

D

150°

Question 10: Palette 10/53

DPT MCQ

Concept: Vector Addition (Parallelogram Law)

If magnitude of resultant of 2N and 3N is 4N find angle between 2N and 3N:

A

$cos^{-1}(1/4)$

B

$cos^{-1}(1/3)$

C

$cos^{-1}(1/2)$

D

$cos^{-1}(3/4)$

Question 11: Palette 11/53

DPT MCQ

Concept: Analytical method of vector addition

Two vectors of magnitude 4N and 6N are acting at an angle $60^{\circ}$ then find:
(i) Magnitude of their resultant vectors
(ii) Angle between resultant vector and 4N (iii) Angle between resultant vector and 6N

A

(i) $2\sqrt{19}N$, (ii) $tan^{-1}(\frac{3\sqrt{3}}{7})$, (iii) $tan^{-1}(\frac{\sqrt{3}}{4})$

B

(i) $\sqrt{52}N$, (ii) $tan^{-1}(\frac{3\sqrt{3}}{7})$, (iii) $tan^{-1}(\frac{\sqrt{3}}{4})$

C

(i) $2\sqrt{19}N$, (ii) $tan^{-1}(\frac{2\sqrt{3}}{7})$, (iii) $tan^{-1}(\frac{\sqrt{3}}{5})$

D

(i) $10N$, (ii) $30^{\circ}$, (iii) $30^{\circ}$

Question 12: Palette 12/53

DPT MCQ

Concept: Range of resultant of two vectors

What can be resultant of two vectors $\vec{A}$ and $\vec{B}$ of magnitude 3 N and 5 N?

A

1 N, 10 N, 12 N

B

2 N, 5 N, 6 N

C

0 N, 1 N, 10 N

D

10 N, 12 N, 0 N

Question 13: Palette 13/53

DPT MCQ

Concept: Resultant of n coplanar vectors of equal magnitude at equal angular separation

If 100 coplanar vectors each having magnitude 10 units are equally inclined with each other then find the magnitude of their resultant?

A

10

B

100

C

1000

Question 14: Palette 14/53

DPT MCQ

Concept: Condition for zero resultant of three vectors (Triangle Inequality)

Which of the following groups of vectors can give a zero resultant (Equilibrium)?

A

2N, 4N, 8N

B

10N, 12N, 23N

C

1N, 2N, 3N

D

1N, 8N, 2N

Question 15: Palette 15/53

DPT MCQ

Concept: Resultant of n coplanar vectors of equal magnitude at equal angular separation

It is a regular hexagon and O is the centre then find $\vec{OA} + \vec{OB} + \vec{OC} + \vec{OD} + \vec{OE} + \vec{OF} = ?$

A

$\vec{0}$

B

$6\vec{a}$

C

$3\vec{a}$

D

$\vec{a}$

Question 16: Palette 16/53

DPT MCQ

Concept: Magnitude of vector addition and subtraction

Find the angle between $\vec{A}$ and $\vec{B}$ if $|\vec{A} - \vec{B}| = |\vec{A} + \vec{B}|$

A

$90^{\circ}$

B

$60^{\circ}$

C

$0^{\circ}$

D

$180^{\circ}$

Question 17: Palette 17/53

DPT MCQ

Concept: A unit vector is a vector that has a magnitude of exactly 1. It represents direction only. The unit vector $\hat{A}$ in the direction of any vector $\vec{A}$ is given by dividing the vector by its magnitude:
$\hat{A} = \frac{\vec{A}}{\vert{}\vec{A}\vert{}}$

The unit vector along $\hat{i}-2\hat{j}$ is:

A

$\frac{\hat{i}-2\hat{j}}{\sqrt{5}}$

B

$\hat{i}+\hat{j}$

C

$\frac{\hat{i}+\hat{j}}{\sqrt{2}}$

D

$\frac{\hat{i}-\hat{j}}{\sqrt{5}}$

Question 18: Palette 18/53

DPT MCQ

Concept: The magnitude of a unit vector is always equal to 1. For any vector $\vec{A} = x\hat{i} + y\hat{j} + z\hat{k}$ to be a unit vector, its magnitude must satisfy the condition:
$\sqrt{x^2 + y^2 + z^2} = 1$

If a unit vector is represented by $0.3\hat{i}-0.4\hat{j}+c\hat{k}$, then the value of c is:

A

$\sqrt{0.75}$

B

$\sqrt{0.25}$

C

$\sqrt{0.01}$

D

$\sqrt{0.39}$

Question 19: Palette 19/53

DPT MCQ

Concept: Work done by a constant force is given by the scalar product (dot product) of the force vector and the displacement vector:
$W = \vec{F} \cdot \vec{S}$
For two vectors $\vec{A} = a_1\hat{i} + a_2\hat{j}$ and $\vec{B} = b_1\hat{i} + b_2\hat{j}$, their dot product is:
$\vec{A} \cdot \vec{B} = a_1b_1 + a_2b_2$

A force $\vec{F}=(2\hat{i}+3\hat{j})\text{ N}$ acts on a body and displaces it by $\vec{S}=(3\hat{i}+4\hat{j})\text{ m}$. The work done $(W=\vec{F}\cdot\vec{S})$ by the force is:

A

10 J

B

12 J

C

18 J

D

25 J

Question 20: Palette 20/53

DPT MCQ

Concept: Work done by a force during a displacement is given by the dot product of the force vector and the displacement vector:
$W = \vec{F} \cdot \vec{S}$
For any two vectors $\vec{A} = a_1\hat{i} + a_2\hat{j}$ and $\vec{B} = b_1\hat{i} + b_2\hat{j}$, their dot product is calculated as:
$\vec{A} \cdot \vec{B} = a_1b_1 + a_2b_2$

A force $\vec{F} = (2\hat{i}+2\hat{j})\text{ N}$ displaces an object through a distance $\vec{S} = (2\hat{i}-3\hat{j})\text{ m}$. The work done ($W=\vec{F}\cdot\vec{S}$) is:

A

-2 J

B

12 J

C

5 J

D

13 J

Question 21: Palette 21/53

DPT MCQ

uestion:

Concept: Work done is the dot product of the force vector and the displacement vector:
$W = \vec{F} \cdot \vec{S}$
The force vector can be determined by multiplying the magnitude of the force by its unit vector direction:
$\vec{F} = F \hat{n} = F \frac{\vec{A}}{\vert{}\vec{A}\vert{}}$
The displacement vector $\vec{S}$ from an initial position $\vec{r}_1$ to a final position $\vec{r}_2$ is:
$\vec{S} = \vec{r}_2 - \vec{r}_1$

A force of $14\text{ N}$ acts on a particle along the vector $(3\hat{i}+2\hat{j}-6\hat{k})$. If the particle displaces from $(0, 0, 0)$ to $(2, 4, -2)$, the work done $(W=\vec{F}\cdot\vec{S})$ by the force on the particle is:

A

10 J

B

52 J

C

-48 J

D

14 J

Question 22: Palette 22/53

DPT MCQ

Concept: Two non-zero vectors $\vec{A}$ and $\vec{B}$ are perpendicular (orthogonal) to each other if and only if their scalar product (dot product) is zero:
$\vec{A} \cdot \vec{B} = 0$
For vectors $\vec{A} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ and $\vec{B} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$, the dot product is:
$\vec{A} \cdot \vec{B} = a_1b_1 + a_2b_2 + a_3b_3$

If a vector $(2\hat{i}+3\hat{j}+8\hat{k})$ is perpendicular to the vector $(4\hat{i}-4\alpha\hat{j}+\alpha\hat{k})$, then the value of $\alpha$ is:

A

-1

B

1/2

C

-1/2

D

2

Question 23: Palette 23/53

DPT MCQ

Concept: For two vectors to be perpendicular, their scalar product (dot product) must be equal to zero:
$\vec{A} \cdot \vec{B} = 0$
Additionally, since $\hat{n}$ is a unit vector, its magnitude must be exactly 1:
$\sqrt{a^2 + b^2} = 1$

If $\hat{n} = a\hat{i}+b\hat{j}$ is a unit vector perpendicular to the vector $(\hat{i}-\hat{j})$, then the value of a and b may be:

A

1, 0

B

-2, 0

C

3, 0

D

$\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}$

Question 24: Palette 24/53

DPT MCQ

Concept: The angle $\theta$ between two vectors $\vec{A}$ and $\vec{B}$ can be found using the definition of the scalar product (dot product):
$\vec{A} \cdot \vec{B} = \vert{}\vec{A}\vert{} \vert{}\vec{B}\vert{} \cos\theta$
Alternatively, if one vector is a positive scalar multiple of another ($\vec{B} = k\vec{A}$ where $k > 0$), the vectors point in the exact same direction, meaning they are parallel, and the angle between them is $0^{\circ}$.

The angle between the two vectors $\vec{A}=3\hat{i}+4\hat{j}+5\hat{k}$ and $\vec{B}=6\hat{i}+8\hat{j}+10\hat{k}$ will be:

A

$0^{\circ}$

B

$180^{\circ}$

C

$90^{\circ}$

D

$45^{\circ}$

Question 25: Palette 25/53

DPT MCQ

Concept: The angle $\theta$ between two vectors $\vec{A}$ and $\vec{B}$ can be determined using the definition of the scalar product (dot product):
$\vec{A} \cdot \vec{B} = \vert{}\vec{A}\vert{} \vert{}\vec{B}\vert{} \cos\theta$
Rearranging the formula gives:
$\cos\theta = \frac{\vec{A} \cdot \vec{B}}{\vert{}\vec{A}\vert{} \vert{}\vec{B}\vert{}}$

The angle between vectors $(\hat{i}+\hat{j})$ and $(\hat{i}+\hat{k})$ is:

A

$90^{\circ}$

B

$180^{\circ}$

C

$0^{\circ}$

D

$60^{\circ}$

Question 26: Palette 26/53

DPT MCQ

Concept: The angle $\theta$ between two vectors $\vec{A}$ and $\vec{B}$ can be determined using the definition of the scalar product (dot product):
$\vec{A} \cdot \vec{B} = \vert{}\vec{A}\vert{} \vert{}\vec{B}\vert{} \cos\theta$
Rearranging the formula gives:
$\cos\theta = \frac{\vec{A} \cdot \vec{B}}{\vert{}\vec{A}\vert{} \vert{}\vec{B}\vert{}}$

The angle between $2\hat{i}+\hat{j}+2\hat{k}$ and $\hat{i}-\hat{j}+\hat{k}$ is:

A

$30^{\circ}$

B

$60^{\circ}$

C

$\cos^{-1}\left(\frac{1}{\sqrt{3}}\right)$

D

$\cos^{-1}\left(\frac{2}{3}\right)$

Question 27: Palette 27/53

DPT MCQ

Concept: The projection of any vector $\vec{A}$ along a specific axis is given by the component of the vector along that axis. For a vector written in component form $\vec{A} = x\hat{i} + y\hat{j} + z\hat{k}$, its projections along the x, y, and z axes are its respective scalar components $x$, $y$, and $z$.
Alternatively, the projection of a vector $\vec{A}$ along the direction of a unit vector $\hat{n}$ is found using the dot product:
$\text{Projection} = \vec{A} \cdot \hat{n}$

What is the projection of $3\hat{i}+4\hat{k}$ on the z-axis?

A

3

B

4

C

5

D

zero

Question 28: Palette 28/53

DPT MCQ

Concept: The scalar projection of a vector $\vec{B}$ on another vector $\vec{A}$ represents the magnitude of the component of $\vec{B}$ that lies in the direction of $\vec{A}$. It is calculated by taking the dot product of vector $\vec{B}$ with the unit vector along $\vec{A}$ (denoted as $\hat{A}$):
$\text{Projection of } \vec{B} \text{ on } \vec{A} = \vec{B} \cdot \hat{A}$
Since the unit vector is defined as $\hat{A} = \frac{\vec{A}}{\vert{}\vec{A}\vert{}}$, the projection can also be written as:
$\text{Projection} = \frac{\vec{A} \cdot \vec{B}}{\vert{}\vec{A}\vert{}}$

What is the projection of vector $\vec{B}$ on vector $\vec{A}$?

A

$\vec{A} \cdot \vec{B}$

B

$\vec{A} \cdot \hat{B}$

C

$\vec{B} \cdot \hat{A}$

D

$\hat{A} \cdot \hat{B}$

Question 29: Palette 29/53

DPT MCQ

Concept: The angle $\theta$ between two vectors $\vec{U}$ and $\vec{V}$ can be found using their scalar product (dot product):
$\vec{U} \cdot \vec{V} = \vert{}\vec{U}\vert{} \vert{}\vec{V}\vert{} \cos\theta$
For any vector $\vec{A}$, the dot product with itself gives its magnitude squared:
$\vec{A} \cdot \vec{A} = \vert{}\vec{A}\vert{}^2$
Since $\vec{A}$ and $\vec{B}$ are perpendicular, their dot product is zero:
$\vec{A} \cdot \vec{B} = 0$

Given that $\vert{}\vec{A}\vert{} = \vert{}\vec{B}\vert{}$ and $\vec{A} \perp \vec{B}$. What is the angle between $(\vec{A}+\vec{B})$ and $(\vec{A}-\vec{B})$?

A

$30^{\circ}$

B

$60^{\circ}$

C

$90^{\circ}$

D

$180^{\circ}$

Question 30: Palette 30/53

DPT MCQ

Concept: A vector can be represented as the product of its magnitude and the unit vector in its direction: $\vec{p} = p \hat{A}$, where $\hat{A} = \frac{\vec{A}}{\vert{}\vec{A}\vert{}}$.

A particle has momentum of magnitude $20\text{ kg}\cdot\text{m/s}$. If the momentum is in the direction of $\vec{A}$, find the momentum in vector form, where $\vec{A} = \hat{i} + \hat{j}$.

A

$10\hat{i} + 10\hat{j}\text{ kg}\cdot\text{m/s}$

B

$10\sqrt{2}\hat{i} + 10\sqrt{2}\hat{j}\text{ kg}\cdot\text{m/s}$

C

$20\hat{i} + 20\hat{j}\text{ kg}\cdot\text{m/s}$

D

$20\sqrt{2}\hat{i} + 20\sqrt{2}\hat{j}\text{ kg}\cdot\text{m/s}$

Question 31: Palette 31/53

DPT MCQ

Concept: The scalar component of a vector $\vec{A}$ along another vector $\vec{B}$ is given by the formula $A \cos \theta = \frac{\vec{A} \cdot \vec{B}}{B}$, where $\vec{A} \cdot \vec{B}$ is the dot product of the two vectors and $B$ is the magnitude of vector $\vec{B}$. The vector component is obtained by multiplying this scalar component by the unit vector in the direction of $\vec{B}$, which is $\hat{B} = \frac{\vec{B}}{B}$. Therefore, the vector component is $\frac{\vec{A} \cdot \vec{B}}{B^2} \vec{B}$.

Find the component of $\vec{A} = 3\hat{i} + 4\hat{j}$ along $\vec{B} = \hat{i} + \hat{j}$.

A

Scalar: $\frac{7}{\sqrt{2}}$, Vector: $\frac{7}{2}(\hat{i} + \hat{j})$

B

Scalar: $\frac{5}{\sqrt{2}}$, Vector: $\frac{5}{2}(\hat{i} + \hat{j})$

C

Scalar: $7$, Vector: $7(\hat{i} + \hat{j})$

D

Scalar: $\frac{7}{2}$, Vector: $\frac{7}{\sqrt{2}}(\hat{i} + \hat{j})$

Question 32: Palette 32/53

DPT MCQ

Concept: The velocity vector $\vec{v}$ of an object is represented as the product of its speed $v$ and the unit vector in its direction of motion $\hat{A}$: $\vec{v} = v\hat{A}$, where $\hat{A} = \frac{\vec{A}}{\vert{}\vec{A}\vert{}}$.

A bird is flying with speed $10\text{ m/s}$ in the direction of a vector $\vec{A} = 3\hat{i} + 4\hat{j}$. Find the velocity of the bird in vector form.

A

$3\hat{i} + 4\hat{j}\text{ m/s}$

B

$6\hat{i} + 8\hat{j}\text{ m/s}$

C

$10\hat{i} + 10\hat{j}\text{ m/s}$

D

$15\hat{i} + 20\hat{j}\text{ m/s}$

Question 33: Palette 33/53

DPT MCQ

Concept: The vector component of $\vec{A}$ parallel to $\vec{B}$ is given by $\vec{A}_{\vert{}\vert{}} = \left(\frac{\vec{A} \cdot \vec{B}}{B^2}\right) \vec{B}$. The component of $\vec{A}$ perpendicular to $\vec{B}$ is given by subtracting the parallel component from the vector itself, which is $\vec{A}_{\perp} = \vec{A} - \vec{A}_{\vert{}\vert{}}$.

If $\vec{A} = \hat{i} + 3\hat{j}$ and $\vec{B} = 3\hat{i} + 4\hat{j}$, find the component of $\vec{A}$ perpendicular to $\vec{B}$.

A

$\frac{-4\hat{i} + 3\hat{j}}{5}$

B

$\frac{-4\hat{i} + 3\hat{j}}{25}$

C

$\frac{-12\hat{i} + 16\hat{j}}{25}$

D

$\frac{-4\hat{i} - 3\hat{j}}{5}$

Question 34: Palette 34/53

DPT MCQ

Concept: Two vectors $\vec{A}$ and $\vec{B}$ are parallel or anti-parallel if their corresponding components are proportional:
$\frac{A_x}{B_x} = \frac{A_y}{B_y} = \frac{A_z}{B_z} = n$
If $n > 0$, the vectors are parallel.
If $n < 0$, the vectors are anti-parallel.

For the vectors $\vec{A} = 3\hat{i} - 4\hat{j} + 5\hat{k}$ and $\vec{B} = -9\hat{i} + 12\hat{j} - 15\hat{k}$, determine the relationship between them.

A

The vectors are parallel

B

The vectors are anti-parallel

C

The vectors are perpendicular

D

The vectors are equal

Question 35: Palette 35/53

DPT MCQ

Concept: The vector component of $\vec{A}$ parallel to $\vec{B}$ is given by $\vec{A}_{\vert{}\vert{}} = \left(\frac{\vec{A} \cdot \vec{B}}{B^2}\right) \vec{B}$. The component of $\vec{A}$ perpendicular to $\vec{B}$ is given by subtracting this parallel component from the original vector: $\vec{A}_{\perp} = \vec{A} - \vec{A}_{\vert{}\vert{}}$.

If $\vec{A} = 4\hat{i} - 2\hat{j}$ and $\vec{B} = 3\hat{i} + 4\hat{j}$, find the component of $\vec{A}$ perpendicular to $\vec{B}$.

A

$\frac{88\hat{i} - 66\hat{j}}{25}$

B

$\frac{88\hat{i} + 66\hat{j}}{25}$

C

$\frac{12\hat{i} + 16\hat{j}}{25}$

D

$\frac{4\hat{i} - 2\hat{j}}{5}$

Question 36: Palette 36/53

DPT MCQ

Concept: The direction cosines of a vector $\vec{A} = A_x\hat{i} + A_y\hat{j} + A_z\hat{k}$ are given by $\cos \alpha = \frac{A_x}{A}$, $\cos \beta = \frac{A_y}{A}$, and $\cos \gamma = \frac{A_z}{A}$, where $A = \sqrt{A_x^2 + A_y^2 + A_z^2}$ is the magnitude of the vector. Here, $\alpha$, $\beta$, and $\gamma$ are the angles made by the vector with the positive x, y, and z axes respectively.

If $\vec{A} = 2\hat{i} + 3\hat{j} + 6\hat{k}$, find its direction cosines.

A

$\cos\alpha = \frac{2}{7}$, $\cos\beta = \frac{3}{7}$, $\cos\gamma = \frac{6}{7}$

B

$\cos\alpha = \frac{2}{\sqrt{11}}$, $\cos\beta = \frac{3}{\sqrt{11}}$, $\cos\gamma = \frac{6}{\sqrt{11}}$

C

$\cos\alpha = \frac{3}{7}$, $\cos\beta = \frac{2}{7}$, $\cos\gamma = \frac{6}{7}$

D

$\cos\alpha = \frac{1}{7}$, $\cos\beta = \frac{2}{7}$, $\cos\gamma = \frac{3}{7}$

Question 37: Palette 37/53

DPT MCQ

Concept: For a vector $\vec{A} = A_x\hat{i} + A_y\hat{j} + A_z\hat{k}$, the magnitude is given by $A = \sqrt{A_x^2 + A_y^2 + A_z^2}$. The direction cosines are defined as $\cos \alpha = \frac{A_x}{A}$, $\cos \beta = \frac{A_y}{A}$, and $\cos \gamma = \frac{A_z}{A}$. The components along the x, y, and z axes are given by $A_x\hat{i}$, $A_y\hat{j}$, and $A_z\hat{k}$ respectively.

If $\vec{A} = 3\hat{i} + 4\hat{j} + 5\hat{k}$, find its direction cosines and components along the x, y, and z axes.

A

Direction cosines: $\cos\alpha = \frac{3}{5\sqrt{2}}$, $\cos\beta = \frac{4}{5\sqrt{2}}$, $\cos\gamma = \frac{1}{\sqrt{2}}$; Components: $3\hat{i}$, $4\hat{j}$, $5\hat{k}$

B

Direction cosines: $\cos\alpha = \frac{3}{5}$, $\cos\beta = \frac{4}{5}$, $\cos\gamma = 1$; Components: $3\hat{i}$, $4\hat{j}$, $5\hat{k}$

C

Direction cosines: $\cos\alpha = \frac{3}{\sqrt{2}}$, $\cos\beta = \frac{4}{\sqrt{2}}$, $\cos\gamma = \frac{5}{\sqrt{2}}$; Components: $3\hat{i}$, $4\hat{j}$, $5\hat{k}$

D

Direction cosines: $\cos\alpha = \frac{3}{5\sqrt{2}}$, $\cos\beta = \frac{2}{5\sqrt{2}}$, $\cos\gamma = \frac{1}{\sqrt{2}}$; Components: $\hat{i}$, $\hat{j}$, $\hat{k}$

Question 38: Palette 38/53

DPT MCQ

Concept: The cross product of two vectors in the xy-plane $\vec{a} = a_x\hat{i} + a_y\hat{j}$ and $\vec{b} = b_x\hat{i} + b_y\hat{j}$ is calculated using the distributive property, keeping in mind the standard unit vector cross products: $\hat{i} \times \hat{i} = 0$, $\hat{j} \times \hat{j} = 0$, $\hat{i} \times \hat{j} = \hat{k}$, and $\hat{j} \times \hat{i} = -\hat{k}$. Thus, $\vec{a} \times \vec{b} = (a_x b_y - a_y b_x)\hat{k}$.

Find the cross product $\vec{a} \times \vec{b}$ if $\vec{a} = 3\hat{i} + 4\hat{j}$ and $\vec{b} = 2\hat{i} + 5\hat{j}$.

A

$7\hat{k}$

B

$-7\hat{k}$

C

$23\hat{k}$

D

$15\hat{k}$

Question 39: Palette 39/53

DPT MCQ

Concept: The cross product of two vectors in the xy-plane $\vec{a} = a_x\hat{i} + a_y\hat{j}$ and $\vec{b} = b_x\hat{i} + b_y\hat{j}$ is determined using the distributive law along with the standard unit vector relationships: $\hat{i} \times \hat{i} = 0$, $\hat{j} \times \hat{j} = 0$, $\hat{i} \times \hat{j} = \hat{k}$, and $\hat{j} \times \hat{i} = -\hat{k}$. The resulting formula is $\vec{a} \times \vec{b} = (a_x b_y - a_y b_x)\hat{k}$.

Find the cross product $\vec{a} \times \vec{b}$ if $\vec{a} = 4\hat{i} + 7\hat{j}$ and $\vec{b} = 2\hat{i} + 3\hat{j}$.

A

$-2\hat{k}$

B

$2\hat{k}$

C

$-26\hat{k}$

D

$26\hat{k}$

Question 40: Palette 40/53

DPT MCQ

Concept: The magnitude of the sum of two vectors $\vec{P}$ and $\vec{Q}$ with an angle $\theta$ between them is given by $\vert{}\vec{P} + \vec{Q}\vert{} = \sqrt{P^2 + Q^2 + 2PQ\cos\theta}$. The magnitude of their difference is given by $\vert{}\vec{P} - \vec{Q}\vert{} = \sqrt{P^2 + Q^2 - 2PQ\cos\theta}$. When the two vectors have equal magnitudes ($P = Q = x$), these formulas simplify using trigonometric identities to $\vert{}\vec{P} + \vec{Q}\vert{} = 2x\cos(\theta/2)$ and $\vert{}\vec{P} - \vec{Q}\vert{} = 2x\sin(\theta/2)$.

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 the angle between $\vec{P}$ and $\vec{Q}$ is:

A

$\cos^{-1}\left(\frac{n^2 - 1}{n^2 + 1}\right)$

B

$\cos^{-1}\left(\frac{n - 1}{n + 1}\right)$

C

$\sin^{-1}\left(\frac{n^2 - 1}{n^2 + 1}\right)$

D

$\tan^{-1}\left(\frac{n^2 - 1}{n^2 + 1}\right)$

Question 41: Palette 41/53

DPT MCQ

Concept: The sum of two vectors $\vec{A} = A_x\hat{i} + A_y\hat{j} + A_z\hat{k}$ and $\vec{B} = B_x\hat{i} + B_y\hat{j} + B_z\hat{k}$ is found by adding their respective components: $\vec{R} = \vec{A} + \vec{B} = (A_x + B_x)\hat{i} + (A_y + B_y)\hat{j} + (A_z + B_z)\hat{k}$. A unit vector $\hat{R}$ in the direction of a vector $\vec{R}$ is calculated by dividing the vector by its magnitude: $\hat{R} = \frac{\vec{R}}{\vert{}\vec{R}\vert{}}$, where $\vert{}\vec{R}\vert{} = \sqrt{R_x^2 + R_y^2 + R_z^2}$.

If $\vec{A} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{B} = 2\hat{i} - \hat{j} + 4\hat{k}$, then the unit vector along $\vec{A} + \vec{B}$ is:

A

$\frac{3\hat{i} + 5\hat{k}}{\sqrt{24}}$

B

$\frac{3\hat{i} - 5\hat{k}}{\sqrt{34}}$

C

$\frac{3\hat{i} + 5\hat{k}}{\sqrt{34}}$

D

None of these

Question 42: Palette 42/53

DPT MCQ

Concept: The cosine of the angle $\theta$ between any two vectors $\vec{X}$ and $\vec{Y}$ is given by the dot product formula $\cos\theta = \frac{\vec{X} \cdot \vec{Y}}{\vert{}\vec{X}\vert{} \vert{}\vec{Y}\vert{}}$. In this problem, we find the resultant vector $\vec{R} = \vec{A} + \vec{B}$ and then determine the cosine of the angle between $\vec{A}$ and $\vec{R}$ using $\cos\theta = \frac{\vec{A} \cdot \vec{R}}{A R}$.

If $\vec{A} = 4\hat{i} + 3\hat{j}$ and $\vec{B} = 3\hat{i} + 4\hat{j}$, then the cosine of the angle between $\vec{A}$ and $\vec{A} + \vec{B}$ is:

A

$\frac{9\sqrt{2}}{5}$

B

$\frac{7}{5\sqrt{2}}$

C

$\frac{5\sqrt{2}}{49}$

D

$\frac{5\sqrt{2}}{28}$

Question 43: Palette 43/53

DPT MCQ

Concept: Basic vector operations including addition, subtraction, scalar (dot) product, and vector (cross) product are performed component-wise. The unit vector cross products follow the rules $\hat{i} \times \hat{i} = 0$, $\hat{j} \times \hat{j} = 0$, and $\hat{i} \times \hat{j} = \hat{k}$, which implies $\hat{j} \times \hat{i} = -\hat{k}$.

Given two vectors $\vec{A} = \hat{i} + \hat{j}$ and $\vec{B} = \hat{i} - \hat{j}$, match the terms in Column-I with Column-II:
Column-I:
(A) $(\vec{A} + \vec{B}) / 2$
(B) $(\vec{A} - \vec{B}) / 2$
(C) $(\vec{A} \cdot \vec{B}) / 2$
(D) $(\vec{A} \times \vec{B}) / 2$
Column-II:
(1) $\hat{i}$
(2) $\hat{j}$
(3) $-\hat{k}$
(4) $0$

A

(A) to (4); (B) to (1); (C) to (2); (D) to (2)

B

(A) to (2); (B) to (4); (C) to (3); (D) to (1)

C

(A) to (3); (B) to (2); (C) to (4); (D) to (1)

D

(A) to (1); (B) to (2); (C) to (4); (D) to (3)

Question 44: Palette 44/53

DPT MCQ

Concept: The scalar component of a vector $\vec{a}$ along another vector $\vec{b}$ is given by the formula $\text{Component} = \frac{\vec{a} \cdot \vec{b}}{\vert{}\vec{b}\vert{}}$, where $\vec{a} \cdot \vec{b}$ is the dot product of the two vectors and $\vert{}\vec{b}\vert{}$ is the magnitude of the vector along which the component is to be found.

The component of vector $\vec{a} = 2\hat{i} + 3\hat{j}$ along the vector $\hat{i} + \hat{j}$ is

A

$\frac{5}{\sqrt{2}}$

B

$10\sqrt{2}$

C

$5\sqrt{2}$

D

5

Question 45: Palette 45/53

DPT MCQ

Concept: The area of a triangle formed by two adjacent vector sides $\vec{A}$ and $\vec{B}$ (along with their resultant completing the triangle) is equal to half the magnitude of their cross product: $\text{Area} = \frac{1}{2}\vert{}\vec{A} \times \vec{B}\vert{}$. The cross product $\vec{A} \times \vec{B}$ is computed using the determinant method.

What is the area of triangle formed by $\vec{A} = 2\hat{i} - 3\hat{j} + 4\hat{k}$ and $\vec{B} = \hat{i} - \hat{k}$ and their resultant?

A

$\sqrt{13.5}$ units

B

$13.5$ units

C

$\sqrt{38.7}$ units

D

$38.7$ units

Question 46: Palette 46/53

DPT MCQ

Concept: Two vectors $\vec{A}$ and $\vec{B}$ are parallel if one can be expressed as a scalar multiple of the other, i.e., $\vec{A} = k\vec{B}$, where $k$ is a scalar constant. For parallel vectors, the angle between them is $0^\circ$, and their cross product is zero ($\vec{A} \times \vec{B} = 0$).

If $\vec{A} = 4\hat{i} + 6\hat{j}$ and $\vec{B} = 2\hat{i} + 3\hat{j}$, then which of the following is correct?

A

$\vec{A} \cdot \vec{B} = 29$

B

$\vec{A} \times \vec{B} = 0$

C

$\frac{\vert{}\vec{B}\vert{}}{\vert{}\vec{A}\vert{}} = \frac{2}{1}$

D

angle between $\vec{A}$ and $\vec{B}$ is $30^\circ$

Question 47: Palette 47/53

DPT MCQ

Concept: Two vectors $\vec{A}$ and $\vec{B}$ are parallel if one vector is a scalar multiple of the other, which means $\vec{A} = k\vec{B}$ where $k$ is a scalar constant. For any two parallel vectors, the angle between them is $0^\circ$, and their cross product is zero because $\sin 0^\circ = 0$.

If $\vec{A} = 4\hat{i} + 6\hat{j}$ and $\vec{B} = 2\hat{i} + 3\hat{j}$, then which of the following is correct?

A

$\vec{A} \cdot \vec{B} = 29$

B

$\vec{A} \times \vec{B} = 0$

C

$\frac{\vert{}\vec{B}\vert{}}{\vert{}\vec{A}\vert{}} = \frac{2}{1}$

D

angle between $\vec{A}$ and $\vec{B}$ is $30^\circ$

Question 48: Palette 48/53

DPT MCQ

Concept: If a vector $\vec{a}$ lies in the plane of two vectors $\vec{b}$ and $\vec{c}$, it can be written as a linear combination of them: $\vec{a} = \lambda\vec{b} + \mu\vec{c}$. Furthermore, if it bisects the angle between $\vec{b}$ and $\vec{c}$, and since the magnitudes of $\vec{b}$ and $\vec{c}$ are equal ($\vert{}\vec{b}\vert{} = \vert{}\vec{c}\vert{} = \sqrt{2}$), the angular bisector must be parallel to either $\vec{b} + \vec{c}$ (internal bisector) or $\vec{b} - \vec{c}$ (external bisector). Therefore, $\vec{a} = k(\vec{b} \pm \vec{c})$ for some scalar $k$.

The vector $\vec{a} = \alpha\hat{i} + 2\hat{j} + \beta\hat{k}$ lies in the plane of the vectors $\vec{b} = \hat{i} + \hat{j}$ and $\vec{c} = \hat{j} + \hat{k}$ and bisects the angle between $\vec{b}$ and $\vec{c}$. Then which one of the following gives possible values of $\alpha$ and $\beta$?

A

$\alpha = 2, \beta = 2$

B

$\alpha = 1, \beta = 2$

C

$\alpha = 2, \beta = 1$

D

$\alpha = 1, \beta = 1$

Question 49: Palette 49/53

DPT MCQ

Concept: Vector addition and unit vectors. A unit vector along the Y-axis is represented as $\hat{j}$ (or $0\hat{i} + 1\hat{j} + 0\hat{k}$).

A vector $\vec{X}$, when added to two vectors $\vec{A} = 3\hat{i} - 5\hat{j} + 7\hat{k}$ and $\vec{B} = 2\hat{i} + 4\hat{j} - 3\hat{k}$ gives a unit vector along Y-axis. Find the vector $\vec{X}$.

A

$-5\hat{i} + 2\hat{j} - 4\hat{k}$

B

$5\hat{i} - 2\hat{j} + 4\hat{k}$

C

$-5\hat{i} - 2\hat{j} + 4\hat{k}$

D

$5\hat{i} + 2\hat{j} - 4\hat{k}$

Question 50: Palette 50/53

DPT MCQ

Concept: Vector Addition and Resultant Magnitude

Two vectors of magnitude 10 N each gives a resultant of magnitude $10\sqrt{3}$ N. Find the angle between both vectors.

A

30°

B

45°

C

60°

D

90°

Question 51: Palette 51/53

DPT MCQ

Concept: System of linear vector equations and solving for unknown vectors by elimination.

Given that $\vec{A} + 2\vec{B} = x_1\hat{i} + y_1\hat{j}$ and $2\vec{A} - \vec{B} = x_2\hat{i} + y_2\hat{j}$, what is $\vec{A}$?

A

$\vec{A} = \frac{1}{5}(x_1 + 2x_2)\hat{i} + \frac{1}{5}(y_1 + 2y_2)\hat{j}$

B

$\vec{A} = \frac{1}{5}(x_1 + 2x_2)\hat{i} - \frac{1}{5}(y_1 + 2y_2)\hat{j}$

C

$\vec{A} = \frac{1}{5}(x_1 + 4x_2)\hat{i} + \frac{1}{5}(y_1 + 2y_2)\hat{j}$

D

$\vec{A} = \frac{1}{5}(x_1 + 4x_2)\hat{i} + \frac{1}{5}(y_1 + 4y_2)\hat{j}$

Question 52: Palette 52/53

DPT MCQ

Concept: Vector subtraction, isosceles triangles formed by rotating a vector of fixed magnitude, and the small-angle approximation $\sin\left(\frac{\Delta\theta}{2}\right) \approx \frac{\Delta\theta}{2}$.

A vector $\vec{A}$ is rotated by a small angle $\Delta\theta$ radians ($\Delta\theta \ll 1$) to get a new vector $\vec{B}$. In that case $\vert{}\vec{B} - \vec{A}\vert{}$ is

A

$0$

B

$\vert{}\vec{A}\vert{} \left(1 - \frac{\Delta\theta^2}{2}\right)$

C

$\vert{}\vec{A}\vert{} \Delta\theta$

D

$\vert{}\vec{B}\vert{} \Delta\theta - \vert{}\vec{A}\vert{}$

Question 53: Palette 53/53

DPT MCQ

Concept: Vector subtraction, isosceles triangles formed by rotating a vector of fixed magnitude, and the small-angle approximation $\sin\left(\frac{\Delta\theta}{2}\right) \approx \frac{\Delta\theta}{2}$.

A vector $\vec{A}$ is rotated by a small angle $\Delta\theta$ radians ($\Delta\theta \ll 1$) to get a new vector $\vec{B}$. In that case $\vert{}\vec{B} - \vec{A}\vert{}$ is

A

$0$

B

$\vert{}\vec{A}\vert{} \left(1 - \frac{\Delta\theta^2}{2}\right)$

C

$\vert{}\vec{A}\vert{} \Delta\theta$

D

$\vert{}\vec{B}\vert{} \Delta\theta - \vert{}\vec{A}\vert{}$