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Vectors Algebra

55 Questions Start DPT Test
All Exams DPT Resnick Haliday Error less
Q1 DPT Resultant MCQ
22 Jul 2026
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
Q2 DPT Resultant MCQ
22 Jul 2026
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
Q3 DPT Resultant MCQ
22 Jul 2026
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
Q4 DPT Resultant MCQ
22 Jul 2026
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
Q5 DPT Resultant MCQ
22 Jul 2026
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
Q6 DPT Resultant MCQ
22 Jul 2026
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°
Q7 DPT Resultant MCQ
22 Jul 2026
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
Q8 DPT Resultant MCQ
22 Jul 2026
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°
Q9 DPT Resultant MCQ
22 Jul 2026
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°
Q10 DPT Resultant MCQ
22 Jul 2026
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.
$cos^{-1}(3/4)$
Q11 DPT Resultant MCQ
22 Jul 2026
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) $30^{\circ}$
Q12 DPT Resultant MCQ
22 Jul 2026
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
Q13 DPT Resultant MCQ
22 Jul 2026
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
Q14 DPT Resultant MCQ
22 Jul 2026
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
Q15 DPT Resultant MCQ
22 Jul 2026
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.
$\vec{a}$
Q16 DPT Resultant MCQ
22 Jul 2026
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.
$180^{\circ}$
Q17 DPT Unit Vector MCQ
22 Jul 2026
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.
$\frac{\hat{i}-\hat{j}}{\sqrt{5}}$
Q18 DPT Unit Vector MCQ
22 Jul 2026
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.
$\sqrt{0.39}$
Q19 DPT Dot Product MCQ
22 Jul 2026
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
Q20 DPT Dot Product MCQ
22 Jul 2026
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
Q21 DPT Dot Product MCQ
22 Jul 2026
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
Q22 DPT Dot Product MCQ
22 Jul 2026
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
Q23 DPT Dot Product MCQ
22 Jul 2026
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.
$\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}$
Q24 DPT Dot Product MCQ
22 Jul 2026
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.
$45^{\circ}$
Q25 DPT Dot Product MCQ
22 Jul 2026
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.
$60^{\circ}$
Q26 DPT Dot Product MCQ
22 Jul 2026
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.
$\cos^{-1}\left(\frac{2}{3}\right)$
Q27 DPT Dot Product MCQ
22 Jul 2026
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
Q28 DPT Dot Product MCQ
22 Jul 2026
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.
$\hat{A} \cdot \hat{B}$
Q29 DPT Dot Product MCQ
22 Jul 2026
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.
$180^{\circ}$
Q30 DPT Unit Vector MCQ
22 Jul 2026
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.
$20\sqrt{2}\hat{i} + 20\sqrt{2}\hat{j}\text{ kg}\cdot\text{m/s}$
Q31 DPT Dot Product MCQ
22 Jul 2026
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: $\frac{7}{\sqrt{2}}(\hat{i} + \hat{j})$
Q32 DPT Unit Vector MCQ
22 Jul 2026
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.
$15\hat{i} + 20\hat{j}\text{ m/s}$
Q33 DPT Dot Product MCQ
22 Jul 2026
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.
$\frac{-4\hat{i} - 3\hat{j}}{5}$
Q34 DPT Dot Product MCQ
22 Jul 2026
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
Q35 DPT Dot Product MCQ
22 Jul 2026
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.
$\frac{4\hat{i} - 2\hat{j}}{5}$
Q36 DPT Dot Product MCQ
22 Jul 2026
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.
$\cos\alpha = \frac{1}{7}$, $\cos\beta = \frac{2}{7}$, $\cos\gamma = \frac{3}{7}$
Q37 DPT Dot Product MCQ
22 Jul 2026
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: $\hat{i}$, $\hat{j}$, $\hat{k}$
Q38 DPT Cross Product MCQ
22 Jul 2026
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.
$15\hat{k}$
Q39 DPT Cross Product MCQ
22 Jul 2026
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.
$26\hat{k}$
Q40 DPT Resultant MCQ
22 Jul 2026
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.
$\tan^{-1}\left(\frac{n^2 - 1}{n^2 + 1}\right)$
Q41 DPT Unit Vector MCQ
22 Jul 2026
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
Q42 DPT Dot Product MCQ
22 Jul 2026
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.
$\frac{5\sqrt{2}}{28}$
Q43 DPT Cross Product MCQ
22 Jul 2026
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)
Q44 DPT Dot Product MCQ
22 Jul 2026
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
Q45 DPT Cross Product MCQ
22 Jul 2026
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
Q46 DPT Cross Product MCQ
22 Jul 2026
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 $30^\circ$
Q47 DPT Cross Product MCQ
22 Jul 2026
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 $30^\circ$
Q48 DPT Dot Product MCQ
22 Jul 2026
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.
$\alpha = 1, \beta = 1$
Q49 DPT Unit Vector MCQ
22 Jul 2026
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.
$5\hat{i} + 2\hat{j} - 4\hat{k}$
Q50 DPT Resultant MCQ
22 Jul 2026
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°