Motion in a Plane (Vectors) MCQs for Class 11 Physics With Answers and Solutions

Motion in a Plane (Vectors) MCQs for Class 11 Physics with Answers and Solutions provides a comprehensive set of multiple-choice questions covering vector addition, subtraction, resolution of vectors, scalar and vector products, unit vectors, and applications of vectors in two-dimensional motion. These practice questions are designed to help students build a strong foundation, improve analytical thinking, and gain confidence for Class 11 Physics examinations as well as competitive exams such as JEE Main, JEE Advanced, and NEET.

Topicwise and Chapterwise Solved MCQs on Motion in a Plane of Topic Vectors for Class 11 Physics

The following are topicwise solved MCQs on Motion in a Plane (Vectors) for Class 11 Physics with detailed answers and solutions.

TOPIC 1: MCQs Based on Topic Scalars and Vectors

Amongst the following quantities, which is not a vector quantity?
(a) Force
(b) Acceleration
(c) Temperature
(d) Velocity

Solution.(c)

Temperature is not a vector quantity because it has magnitude only. However, force, acceleration and velocity have both a magnitude and a direction. So, these are vectors in nature.

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In order to describe the motion in two or three dimensions, we use
(a) positive sign
(b) vectors
(c) negative sign
(d) Both (b) and (c)

Solution.(b)

In order to describe two-dimensional or three-dimensional motions, we use vectors. However, direction of the motion of an object along a straight line is shown by positive and negative signs.

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If length and breadth of a rectangle are 1.0 m and 0.5 m respectively, then its perimeter will be a
(a) free vector
(b) scalar quantity
(c) localised vector
(d) Neither (a) nor (b)

Solution.(b)

The perimeter of the rectangle would be the sum of the lengths of the four sides, i.e.,

1.0 m + 0.5 m + 1.0 m + 0.5 m = 3.0 m

Since length of each side is a scalar, thus the perimeter is also a scalar.

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Set of vectors $\overrightarrow{A}$ and $\overrightarrow{B}$, $\overrightarrow{P}$ and $\overrightarrow{Q}$ are as shown below:
Length of $\overrightarrow{A}$ and $\overrightarrow{B}$ is equal, similarly length of $\overrightarrow{P}$ and $\overrightarrow{Q}$ is equal. Then, the vectors which are equal, are
(a) $\overrightarrow{A}$ and $\overrightarrow{P}$
(b) $\overrightarrow{P}$ and $\overrightarrow{Q}$
(c) $\overrightarrow{A}$ and $\overrightarrow{B}$
(d) $\overrightarrow{B}$ and $\overrightarrow{Q}$

Solution.(c)

Two vectors are said to be equal, if and only if they have the same magnitude and direction. Among the given vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ are equal vectors as they have same magnitude (length) and direction. However, $\overrightarrow{P}$ and $\overrightarrow{Q}$ are not equal even though they are of same magnitude because their directions are different.

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$|\lambda \overrightarrow{A}|=\lambda |\overrightarrow{A}|$, if
(a) λ > 0
(b) λ < 0
(c) λ = 0
(d) λ ≠ 0

Solution.(a)

$|\lambda \overrightarrow{A}|=\lambda |\overrightarrow{A}|$, if λ, as multiplication of vector $\overrightarrow{A}$ with a positive number λ gives a vector whose magnitude is changed by the factor λ but the direction is same as that of $\overrightarrow{A}$.

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If a vector is multiplied by a negative number, we get a vector whose
(a) magnitude and direction both are changed
(b) only direction is changed
(c) only magnitude is changed
(d) only direction is reversed

Solution.(d)

Multiplying a vector $\overrightarrow{A}$ with a negative number $\lambda$ gives a vector whose magnitude is changed by the factor $|\lambda |$ but direction is reversed.

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Choose the correct option regarding the given figure:
(a) $\overrightarrow{B}=\overrightarrow{A}$
(b) $\overrightarrow{B}=-\overrightarrow{A}$
(c) $1.5\overrightarrow{A}=\overrightarrow{B}$
(d) $|\overrightarrow{B}|\ne |\overrightarrow{A}|$

Solution.(d)

$\overrightarrow{B}=-1.5\overrightarrow{A}$. So when $\overrightarrow{A}$ is multiplied by $-1.5$, then its direction gets reversed and magnitude would be $1.5$ times $A$.

Thus, $|\overrightarrow{B}|\ne |\overrightarrow{A}|$.

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$\overrightarrow{A}$ and $\overrightarrow{B}$ are two inclined vectors and $\overrightarrow{R}$ is their sum. Choose the correct figure for the given description:
(a) $\overrightarrow{R}=\overrightarrow{A}+\overrightarrow{B}$
(b) $\overrightarrow{R}=\overrightarrow{A}+\overrightarrow{B}$
(c) $\overrightarrow{R}=\overrightarrow{A}+\overrightarrow{B}$
(d) $\overrightarrow{R}=\overrightarrow{A}+\overrightarrow{B}$

Solution.(d)

Vectors by definition obey the triangle law of addition. According to which, if vector $\overrightarrow{B}$ is placed with its tail at the head of vector $\overrightarrow{A}$, then when we join the tail of $\overrightarrow{A}$ to the head of $\overrightarrow{B}$, the line represents a vector $\overrightarrow{R}$, i.e., the sum of the vectors $\overrightarrow{A}$ and $\overrightarrow{B}$. Thus, figure given in option (d) is correct.

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Among the following properties regarding null vector which is incorrect?
(a) $\overrightarrow{A}+\overrightarrow{0}=\overrightarrow{A}$
(b) $\lambda \overrightarrow{0}=\lambda$
(c) $0\overrightarrow{A}=\overrightarrow{0}$
(d) $\overrightarrow{A}-\overrightarrow{A}=\overrightarrow{0}$

Solution.(b)

Null vector $\overrightarrow{0}$ is a vector whose magnitude is zero and its direction cannot be specified. So, it means $|\overrightarrow{0}|=0$.

Thus, $\lambda \overrightarrow{0}=\overrightarrow{0}$. Hence, property given in option (b) is incorrect.

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Suppose an object is at point P at time t, moves to P’ and then comes back to P. Then, displacement is a
(a) unit vector
(b) null vector
(c) scalar
(d) None of these

Solution. (b)

Since in the given case, the initial and final positions coincide, so the displacement will be zero. Thus, it is a null vector.

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Find the correct option about vector subtraction :
(a) $\overrightarrow{A}-\overrightarrow{B}=\overrightarrow{A}+\overrightarrow{B}$
(b) $\overrightarrow{A}+\overrightarrow{B}=\overrightarrow{B}-\overrightarrow{A}$
(c) $\overrightarrow{A}-\overrightarrow{B}=\overrightarrow{A}+(-\overrightarrow{B})$
(d) None of the above

Solution.(c)

To subtract $\overrightarrow{B}$ from $\overrightarrow{A}$, we can add $-\overrightarrow{B}$ and $\overrightarrow{A}$.

So, $\overrightarrow{A}+(-\overrightarrow{B})=\overrightarrow{A}-\overrightarrow{B}={\overrightarrow{R}}_2$.

Hence, option (c) is correct about vector subtraction.

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TOPIC 2: MCQs Based on Topic Resolution of Vectors

If $\overrightarrow{A}$ is a vector with magnitude A, then the unit vector $\hat{n}$ in the direction of vector $\overrightarrow{A}$ is
(a) $A\overrightarrow{A}$
(b) $\overrightarrow{A}\cdot \overrightarrow{A}$
(c) $\overrightarrow{A}\times \overrightarrow{A}$
(d) ${\displaystyle \frac{\overrightarrow{A}}{|\overrightarrow{A}|}}$

Solution(d) In general, a vector $\overrightarrow{A}$ can be written as :

$\overrightarrow{A}=|\overrightarrow{A}|\hat{n}$

where $\hat{n}$ is a unit vector along $\overrightarrow{A}$. If $\hat{A}$ is a unit vector along $\overrightarrow{A}$, then we can write :

$\hat{A}={\displaystyle \frac{\overrightarrow{A}}{|\overrightarrow{A}|}}$

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Unit vector of $4\hat{i}-3\hat{j}+\hat{k}$ is
(a) $\hat{i}-\hat{j}+\hat{k}$
(b) $\sqrt{26}\hat{i}-\sqrt{26}\hat{j}+\sqrt{26}\hat{k}$
(c) ${\displaystyle \frac{4\hat{i}-3\hat{j}+\hat{k}}{\sqrt{26}}}$
(d) $5\hat{i}-4\hat{j}+5\hat{k}$

Solution.(c)

Given, $\overrightarrow{A}=4\hat{i}-3\hat{j}+\hat{k}$

$|\overrightarrow{A}|=\sqrt{A_x^2+A_y^2+A_z^2}$

$|\overrightarrow{A}|=\sqrt{(4{)}^2+(-3{)}^2+(1{)}^2}=\sqrt{26}$

Unit vector :

$\hat{A}={\displaystyle \frac{\overrightarrow{A}}{|\overrightarrow{A}|}}={\displaystyle \frac{4\hat{i}-3\hat{j}+\hat{k}}{\sqrt{26}}}$

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Consider a vector $\overrightarrow{A}$ that lies in xy-plane. If A_x and A_y are the magnitudes of its x and y-components respectively, then the correct representation of $\overrightarrow{A}$ can be given by
(a) $A\sin \theta \hat{i}+A\cos \theta \hat{j}$
(b) $A\sin \theta \hat{j}-A\cos \theta \hat{i}$
(c) $A\cos \theta \hat{i}+A\sin \theta \hat{j}$
(d) None of these

Solution.(c)

Vector along X-axis (x-component) :

$=A_x\hat{i}=|\overrightarrow{A}|\cos \theta \hat{i}=A\cos \theta \hat{i}$

Vector along Y-axis ($y$-component) :

$=A_y\hat{j}=|\overrightarrow{A}|\sin \theta \hat{j}=A\sin \theta \hat{j}$

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Magnitude of a vector $\overrightarrow{Q}$ is 5 and magnitude of its y-component is 4. So, the magnitude of the x-component of this vector is
(a) 8
(b) 3
(c) 6
(d) 8

Solution.(b)

Given, $|\overrightarrow{Q}|=5$, Q_y = 4, Q_x = ?

$|\overrightarrow{Q}|=\sqrt{Q_x^2+Q_y^2}$

$|\overrightarrow{Q}{|}^2=Q_x^2+Q_y^2$

Substituting the given values, we get :

$(5{)}^2=Q_x^2+4^2⟹Q_x=\sqrt{25-16}=\sqrt{9}=3$

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A vector is inclined at an angle 60° to the horizontal. If its rectangular component in the horizontal direction is 50 N, then its magnitude in the vertical direction is
(a) 25 N
(b) 75 N
(c) 87 N
(d) 100 N

Solution.(c)

Given vector angle θ = 60°. Then,

$\tan \theta ={\displaystyle \frac{A_y}{A_x}}⟹A_y=A_x\tan \theta$

$A_y=50\tan {60}^{\circ }=50\times \sqrt{3}$

$A_y=86.6\approx 87\text{ N}(\because \sqrt{3}=1.732)$

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Three vectors are given as $\overrightarrow{P}=3\hat{i}-4\hat{j}$, $\overrightarrow{Q}=6\hat{i}-8\hat{j}$ and $\overrightarrow{R}=(3/4)\hat{i}-\hat{j}$
then which of the following is correct?
(a) $\overrightarrow{P}$, $\overrightarrow{Q}$ and $\overrightarrow{R}$ are equal vectors
(b) $\overrightarrow{P}$ and $\overrightarrow{Q}$ are parallel but $\overrightarrow{R}$ is not parallel
(c) $\overrightarrow{P}$, $\overrightarrow{Q}$ and $\overrightarrow{R}$ are parallel
(d) None of the above

Solution. (c)

Given, $\overrightarrow{P}=3\hat{i}-4\hat{j}$

$\overrightarrow{Q}=6\hat{i}-8\hat{j}=2(3\hat{i}-4\hat{j})=2\overrightarrow{P}$

$\overrightarrow{R}={\displaystyle \frac{3}{4}}\hat{i}-\hat{j}={\displaystyle \frac{1}{4}}(3\hat{i}-4\hat{j})={\displaystyle \frac{\overrightarrow{P}}{4}}$

So, $\overrightarrow{P}$, $\overrightarrow{Q}$ and $\overrightarrow{R}$ are parallel with unequal magnitude. Thus, they are not equal vectors.

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Two vectors $\overrightarrow{P}$ and $\overrightarrow{Q}$ are inclined at an angle θ and $\overrightarrow{R}$ is their resultant. Keeping the magnitude and the angle of the vectors same, if the direction of $\overrightarrow{P}$ and $\overrightarrow{Q}$ is interchanged, then there is a change in which of the following with regard to $\overrightarrow{R}$?
(a) Magnitude
(b) Direction
(c) Both magnitude and direction
(d) None of the above

Solution.(b)
Since the magnitude and angle between the vectors is unchanged, so the magnitude of the resultant $\overrightarrow{R}$ will be same. However, the direction of $\overrightarrow{R}$ will get changed.

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Consider vectors $\overrightarrow{a}$, $\overrightarrow{b}$ and $\overrightarrow{c}$ as:
$\overrightarrow{a}=a_x\hat{i}+a_y\hat{j}+a_z\hat{k}$
$\overrightarrow{b}=b_x\hat{i}+b_y\hat{j}+b_z\hat{k}$
$\overrightarrow{c}=c_x\hat{i}+c_y\hat{j}+c_z\hat{k}$
Then, for a vector $\overrightarrow{T}=\overrightarrow{a}+\overrightarrow{b}-\overrightarrow{c}$, its y-component has the form
(a) $a_y+b_y+c_y$
(b) $-a_y+b_y-c_y$
(c) $a_y+b_y-c_y$
(d) $a_y-b_y+c_y$

Solution.(c)

$\overrightarrow{T}=\overrightarrow{a}+\overrightarrow{b}-\overrightarrow{c}$

$\overrightarrow{T}=(a_x\hat{i}+a_y\hat{j}+a_z\hat{k})+(b_x\hat{i}+b_y\hat{j}+b_z\hat{k})-(c_x\hat{i}+c_y\hat{j}+c_z\hat{k})$

$\overrightarrow{T}=(a_x+b_x-c_x)\hat{i}+(a_y+b_y-c_y)\hat{j}+(a_z+b_z-c_z)\hat{k}\text{— (i)}$

As $\overrightarrow{T}=T_x\hat{i}+T_y\hat{j}+T_z\hat{k}$ — (ii),

on comparing Eqs. (i) and (ii), we get the $y$-component of $\overrightarrow{T}$:

$T_y=a_y+b_y-c_y$

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Unit vector in the direction of the resultant of vectors $\overrightarrow{A}=-3\hat{i}-2\hat{j}-3\hat{k}$ and $\overrightarrow{B}=2\hat{i}+4\hat{j}+6\hat{k}$ is
(a) ${\displaystyle \frac{-3\hat{i}+2\hat{j}-3\hat{k}}{\sqrt{14}}}$
(b) $-\hat{i}+2\hat{j}+3\hat{k}$
(c) ${\displaystyle \frac{-\hat{i}+2\hat{j}+3\hat{k}}{\sqrt{14}}}$
(d) $-2\hat{i}-4\hat{j}+8\hat{k}$

Solution.(c) Resultant vector of $\overrightarrow{A}$ and $\overrightarrow{B}$ is :

$\overrightarrow{R}=\overrightarrow{A}+\overrightarrow{B}$

$\overrightarrow{R}=(-3\hat{i}-2\hat{j}-3\hat{k})+(2\hat{i}+4\hat{j}+6\hat{k})$

$\overrightarrow{R}=-\hat{i}+2\hat{j}+3\hat{k}$

$|\overrightarrow{R}|=\sqrt{(-1{)}^2+(2{)}^2+(3{)}^2}$

$|\overrightarrow{R}|=\sqrt{1+4+9}=\sqrt{14}$

Unit vector in the direction of $\overrightarrow{R}$ is :

$\hat{R}={\displaystyle \frac{\overrightarrow{R}}{|\overrightarrow{R}|}}$

$\hat{R}={\displaystyle \frac{-\hat{i}+2\hat{j}+3\hat{k}}{\sqrt{14}}}$

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Two forces $\overrightarrow{P}$ and $\overrightarrow{Q}$ of magnitude 2F and 3F, respectively, are at an angle θ with each other. If the force $\overrightarrow{Q}$ is doubled, then their resultant also gets doubled. Then, the angle θ is
(a) 60°
(b) 120°
(c) 30°
(d) 90°

[JEE Main 2019]

Solution.(b)

Resultant force F_r of any two forces F₁ (i.e., P) and F₂ (i.e., Q) with an angle θ between them can be given by vector addition as:

$F_r^2=F_1^2+F_2^2+2F_1{F}_2\cos \theta \text{— (i)}$

In first case $F_1=2F$ and $F_2=3F$ :

$F_r^2=(2F{)}^2+(3F{)}^2+2(2F)(3F)\cos \theta$

$F_r^2=13F^2+12F^2\cos \theta \text{— (ii)}$

In second case $F_1=2F$ and $F_2=6F$ (Force $Q$ gets doubled), and $F_r^{\prime }=2F_r$ (Given). By putting these values in Eq. (i), we get:

$(2F_r{)}^2=(2F{)}^2+(6F{)}^2+2(2F)(6F)\cos \theta$

$4F_r^2=40F^2+24F^2\cos \theta \text{— (iii)}$

Multiplying Eq. (ii) by 4 and comparing with Eq. (iii), we get :

$4(13F^2+12F^2\cos \theta )=40F^2+24F^2\cos \theta$

$52F^2+48F^2\cos \theta =40F^2+24F^2\cos \theta$

$12F^2+24F^2\cos \theta =0⟹\cos \theta =-{\displaystyle \frac{12}{24}}=-{\displaystyle \frac{1}{2}}$

$\theta ={120}^{\circ }(\because \cos {120}^{\circ }=-1/2)$

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It is found that $|\overrightarrow{A}+\overrightarrow{B}|=|\overrightarrow{A}|$. This necessarily implies
(a) $|\overrightarrow{B}|=0$
(b) $\overrightarrow{A},\overrightarrow{B}$ are parallel
(c) $\overrightarrow{A},\overrightarrow{B}$ are perpendicular
(d) $\overrightarrow{A}\cdot \overrightarrow{B}\le 0$

Solution.(a)

Given that

$|\overrightarrow{A}+\overrightarrow{B}|=|\overrightarrow{A}|$

$|\overrightarrow{A}+\overrightarrow{B}{|}^2=|\overrightarrow{A}{|}^2$

$|\overrightarrow{A}{|}^2+|\overrightarrow{B}{|}^2+2|\overrightarrow{A}||\overrightarrow{B}|\cos \theta =|\overrightarrow{A}{|}^2$

where $\theta$ is the angle between $\overrightarrow{A}$ and $\overrightarrow{B}$.

$|\overrightarrow{B}|(|\overrightarrow{B}|+2|\overrightarrow{A}|\cos \theta )=0$

$⟹|\overrightarrow{B}|=0\text{or}|\overrightarrow{B}|+2|\overrightarrow{A}|\cos \theta =0$

$⟹\cos \theta =-{\displaystyle \frac{|\overrightarrow{B}|}{2|\overrightarrow{A}|}}\text{— (i)}$

If $\overrightarrow{A}$ and $\overrightarrow{B}$ are anti-parallel, then $\theta ={180}^{\circ }$. Hence, from Eq. (i):

$\cos {180}^{\circ }=-1=-{\displaystyle \frac{|\overrightarrow{B}|}{2|\overrightarrow{A}|}}⟹|\overrightarrow{B}|=2|\overrightarrow{A}|$

Hence, the given condition can only be implied if either $\overrightarrow{B}=\overrightarrow{0}$ or $\overrightarrow{A}$ and $\overrightarrow{B}$ are anti-parallel provided $B=2A$.

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Find the value of difference of unit vectors $\hat{A}$ and $\hat{B}$ whose angle of intersection is θ.
(a) 2 sin θ/2
(b) 2 cos θ/2
(c) sin θ/2
(d) cos θ/2

Solution.(a)

Difference of unit vectors $\hat{A}$ and $\hat{B}$ can be given as :

$|(\hat{A}-\hat{B}){|}^2=A^2+B^2-2AB\cos \theta$

$|(\hat{A}-\hat{B}){|}^2=1+1-2\cos \theta$

$[\because \hat{A}\text{ and }\hat{B}\text{ are unit vectors}]$

$|(\hat{A}-\hat{B}){|}^2=2-2\cos \theta =2(1-\cos \theta )$

$|(\hat{A}-\hat{B}){|}^2=2\times 2{\sin }^2(\theta /2)=4{\sin }^2(\theta /2)$

$|\hat{A}-\hat{B}|=2\sin (\theta /2)$

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Given, $|\overrightarrow{A}+\overrightarrow{B}|=P$ and $|\overrightarrow{A}-\overrightarrow{B}|=Q$. The value of $P^2+Q^2$ is
(a) $2(A^2+B^2)$
(b) $A^2-B^2$
(c) $A^2+B^2$
(d) $2(A^2-B^2)$

Solution.(a) Given,

$|\overrightarrow{A}+\overrightarrow{B}|=P$

$|\overrightarrow{A}+\overrightarrow{B}{|}^2=P^2$

$P^2=A^2+B^2+2AB\cos \theta \text{— (i)}$

Similarly, for

$|\overrightarrow{A}-\overrightarrow{B}|=Q$

$Q^2=A^2+B^2-2AB\cos \theta \text{— (ii)}$

Adding Eq. (i) and Eq. (ii):

$P^2+Q^2=2(A^2+B^2)$

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For two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$, $|\overrightarrow{A}+\overrightarrow{B}|=|\overrightarrow{A}-\overrightarrow{B}|$ is always true, when
(a) $|\overrightarrow{A}|=|\overrightarrow{B}|\ne 0$
(b) $|\overrightarrow{A}|=|\overrightarrow{B}|\ne 0$ and $\overrightarrow{A}$ and $\overrightarrow{B}$ are parallel or anti-parallel
(c) when either $|\overrightarrow{A}|$ or $|\overrightarrow{B}|$ is zero
(d) None of the above

Solution.(c)

Given,

$|A+B|=|A-B|$

$\sqrt{|A{|}^2+|B{|}^2+2|A||B|\cos \theta }=\sqrt{|A{|}^2+|B{|}^2-2|A||B|\cos \theta }$

$|A{|}^2+|B{|}^2+2|A||B|\cos \theta =|A{|}^2+|B{|}^2-2|A||B|\cos \theta$

$4|A||B|\cos \theta =0$

$|A||B|\cos \theta =0$

$A=0\text{ or }|B|=0\text{ or }\cos \theta =0\Rightarrow \theta ={90}^{\circ }$

Thus, $|A+B|=|A-B|$ is always true, when either $|A|$ or $|B|$ is zero or $A$ and $B$ are perpendicular to each other.

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Two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ have equal magnitudes. The magnitude of $(\overrightarrow{A}+\overrightarrow{B})$ is $n$ times the magnitude of $(\overrightarrow{A}-\overrightarrow{B})$. The angle between $\overrightarrow{A}$ and $\overrightarrow{B}$ is :
(a) ${\sin }^{-1}\left({\displaystyle \frac{n^2-1}{n^2+1}}\right)$
(b) ${\sin }^{-1}\left({\displaystyle \frac{n-1}{n+1}}\right)$
(c) ${\cos }^{-1}\left({\displaystyle \frac{n^2-1}{n^2+1}}\right)$
(d) ${\cos }^{-1}\left({\displaystyle \frac{n-1}{n+1}}\right)$

Solution.(c)

Given,

$|\overrightarrow{A}|=|\overrightarrow{B}|\text{ or }A=B\text{….(i)}$

Let magnitude of $|\overrightarrow{A}+\overrightarrow{B}|$ is $R$ and for $|\overrightarrow{A}-\overrightarrow{B}|$ is $R^{\prime }$.

Now,

$|\overrightarrow{R}|=|\overrightarrow{A}+\overrightarrow{B}|$

$R^2=A^2+B^2+2AB\cos \theta$

$R^2=2A^2+2A^2\cos \theta \text{….(ii) [using Eq. (i)]}$

And,

$|\overrightarrow{R}|=|\overrightarrow{A}-\overrightarrow{B}|$

${R^{\prime }}^2=A^2+B^2-2AB\cos \theta$

${R^{\prime }}^2=2A^2-2A^2\cos \theta \text{….(iii) [using Eq. (i)]}$

Given, $R=n{R}^{\prime }$ or ${\left({\displaystyle \frac{R}{R^{\prime }}}\right)}^2=n^2$

Dividing Eq. (ii) with Eq. (iii), we get

${\displaystyle \frac{n^2}{1}}={\displaystyle \frac{1+\cos \theta }{1-\cos \theta }}$

${\displaystyle \frac{n^2-1}{n^2+1}}={\displaystyle \frac{(1+\cos \theta )-(1-\cos \theta )}{(1+\cos \theta )+(1-\cos \theta )}}$

${\displaystyle \frac{n^2-1}{n^2+1}}={\displaystyle \frac{2\cos \theta }{2}}=\cos \theta$

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

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Rain is falling vertically with a speed of 35 m/s. Wind starts blowing after sometime with a speed of 12 m/s in east to west direction. In which direction from vertical should a boy waiting at a bus stop hold his umbrella?
(a) tan^-1(0.45), west
(b) tan^-1(0.343), west
(c) tan^-1(0.343), east
(d) tan^-1(0.24), east

Solution.(c)

The velocity of the rain and wind are represented by vectors $v_r$ and $v_w$.

Using the rule of vector addition, we see that the resultant of $v_r$ and $v_w$ is R.

The magnitude of R is

$|R|=\sqrt{v_r^2+v_w^2}$

$|R|=\sqrt{{35}^2+{12}^2}=37{\text{ ms}}^{-1}$

The direction that $R$ makes with the vertical is given by

$\tan \theta ={\displaystyle \frac{v_w}{v_r}}$

$\tan \theta ={\displaystyle \frac{12}{35}}=0.343\text{ or }\theta ={\tan }^{-1}(0.343)$

Therefore, the boy should hold his umbrella in the vertical plane at an angle of about ${\tan }^{-1}(0.343)$ with the vertical towards the east.

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Topic -3 : MCQs Based on Motion in a Plane

Position vector $\overrightarrow{r}$ of a particle P located in a plane with reference to the origin of an xy-plane as shown in the figure below is given by:
(a) $2\hat{i}+4\hat{j}$
(b) $4\hat{i}+2\hat{j}$
(c) $6\hat{k}$
(d) $\hat{i}+\hat{j}+4\hat{k}$

Solution.(a)

Position vector r of an object in xy-plane at point P with its components along X and Y-axes as x and y, respectively is given as $r=x\hat{i}+y\hat{j}$.

Given, x = 2 units and y = 4 units. So, position vector at P will be given as

$r=2\hat{i}+4\hat{j}$.

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Suppose a particle moves along a curve and the positions of the particle are represented by P at t and P’ at t’, where coordinates of P are (2, 3) and P’ are (5, 6). The net displacement will be :
(a) zero
(b) $7\hat{i}+9\hat{j}$
(c) $10\hat{i}+18\hat{j}$
(d) $3\hat{i}+3\hat{j}$

Solution.(d)

Position vector of the particle at P,

$r=2\hat{i}+3\hat{j}$

Position vector of the particle at P’,

$r^{\prime }=5\hat{i}+6\hat{j}$

Displacement of the particle is $\mathrm{\Delta }r=r^{\prime }-r$

$\mathrm{\Delta }r=(5\hat{i}+6\hat{j})-(2\hat{i}+3\hat{j})$

$\mathrm{\Delta }r=(5-2)\hat{i}+(6-3)\hat{j}=3\hat{i}+3\hat{j}$

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A particle is moving such that its position coordinates (x, y) are (2 m, 3 m) at t = 0 seconds, (6 m, 7 m) at time 2 seconds and (13 m, 14 m) at time t = 5 seconds. The average velocity vector $({\overrightarrow{v}}_{av})$ from t = 0 seconds to t = 5 seconds is:
(a) ${\displaystyle \frac{1}{5}}(13\hat{i}+14\hat{j})$
(b) ${\displaystyle \frac{11}{5}}(\hat{i}+\hat{j})$
(c) $2(\hat{i}+\hat{j})$
(d) ${\displaystyle \frac{7}{3}}(\hat{i}+\hat{j})$

Solution.(b)

Position vector of the particle at :

$t=0\text{ s}$, $r_{0s}=2\hat{i}+3\hat{j}$

$t=2\text{ s}$, $r_{2s}=6\hat{i}+7\hat{j}$

$t=5\text{ s}$, $r_{5s}=13\hat{i}+14\hat{j}$

Displacement in $t=0\text{ s}$ to $t=5\text{ s}$:

$\mathrm{\Delta }r=r_{5s}-r_{0s}=(13-2)\hat{i}+(14-3)\hat{j}$

$\mathrm{\Delta }r=11\hat{i}+11\hat{j}$

$\text{Average velocity, }\overline{v}={\displaystyle \frac{\mathrm{\Delta }r}{\mathrm{\Delta }t}}$

$\overline{v}={\displaystyle \frac{11\hat{i}+11\hat{j}}{5}}={\displaystyle \frac{11}{5}}(\hat{i}+\hat{j})$

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The position of a particle is given by $\overrightarrow{r}=3t\hat{i}+2t^2\hat{j}+5\hat{k}$. Then the direction of $\overrightarrow{v}(t)$ at t = 1 seconds is:
(a) 45° with X-axis
(b) 63° with Y-axis
(c) 30° with Y-axis
(d) 53° with X-axis

Solution.(d)

Given,

$\overrightarrow{r}=3t\hat{i}+2t^2\hat{j}+5\hat{k}$

At t = 1 seconds,

$\overrightarrow{v(t)}={\displaystyle \frac{dr}{dt}}={\displaystyle \frac{d}{dt}}(3t\hat{i}+2t^2\hat{j}+5\hat{k})=4t\hat{j}$

$v=3\hat{i}+4\hat{j}$

Thus, its direction is

$\theta ={\tan }^{-1}\left({\displaystyle \frac{v_y}{v_x}}\right)$

$\theta ={\tan }^{-1}\left({\displaystyle \frac{4}{3}}\right)$

θ = 53° with X-axis

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The x and y-coordinates of the particle at any time are x = 5t – 2t² and y = 10t respectively, where x and y are in metres and t in seconds. The acceleration of the particle at t = 2 seconds is:
(a) 0
(b) 5 m/s²
(c) -4 m/s²
(d) -8 m/s²

[NEET 2017]

Solution.(c)

Given,

x = 5t – 2t²

$\text{Velocity of the particle, }{v}_x={\displaystyle \frac{dx}{dt}}$

$v_x={\displaystyle \frac{d}{dt}}(5t-2t^2)=5-4t$

$\text{Acceleration, }{a}_x={\displaystyle \frac{d{v}_x}{dt}}=-4{\text{ ms}}^{-2}$

Also,

y = 10 t

$\text{Velocity, }{v}_y={\displaystyle \frac{dy}{dt}}={\displaystyle \frac{d}{dt}}(10t)=10$

$\text{Acceleration, }{a}_y={\displaystyle \frac{d{v}_y}{dt}}=0$

Net acceleration of the particle,

$a_{\text{net}}=a_x\hat{i}+a_y\hat{j}=(-4{\text{ ms}}^{-2})\hat{i}+0$

$a_{\text{net}}=-4\hat{i}{\text{ ms}}^{-2}$

$|a_{\text{net}}|=4{\text{ ms}}^{-2}$

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The position vector of a particle changes with time according to the relation $\overrightarrow{r}(t)=15t^2\hat{i}+(4-20t^2)\hat{j}$. What is the magnitude of the acceleration (in m/s²) at t = 1 seconds ?
(a) 50
(b) 100
(c) 25
(d) 40

[JEE Main 2019]

Solution.(a)

Position vector of particle is given as

$r=15t^2\hat{i}+(4-20t^2)\hat{j}$

$\text{Velocity of particle is }v={\displaystyle \frac{dr}{dt}}$

$v={\displaystyle \frac{d}{dt}}[15t^2\hat{i}+(4-20t^2)\hat{j}]$

$v=30t\hat{i}-40t\hat{j}$

$\text{Acceleration of particle is }a={\displaystyle \frac{d}{dt}}(v)$

$a={\displaystyle \frac{d}{dt}}(30t\hat{i}-40t\hat{j})$

$a=30\hat{i}-40\hat{j}$

So, magnitude of acceleration at t = 1 seconds is

$|a{|}_{t=1s}=\sqrt{a_x^2+a_y^2}=\sqrt{{30}^2+{40}^2}=50{\text{ ms}}^{-2}$

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In a three-dimensional system, the position coordinates of a particle (in motion) are given below:
$x=a\cos \omega t$, $y=a\sin \omega t$, $z=a\omega t$
The velocity of the particle will be:
(a) $\sqrt{2}a\omega$
(b) $2a\omega$
(c) $a\omega$
(d) $\sqrt{3}a\omega$

[JEE Main 2019]

Solution.(a)

Given that the position coordinates of a particle:

$x=a\cos \omega t,y=a\sin \omega t,z=a\omega t$

So, the position vector of the particle is

$\hat{r}=x\hat{i}+y\hat{j}+z\hat{k}$

$\hat{r}=a\cos \omega t\hat{i}+a\sin \omega t\hat{j}+a\omega t\hat{k}$

$\hat{r}=a[\cos \omega t\hat{i}+\sin \omega t\hat{j}+\omega t\hat{k}]$

Therefore, the velocity of the particle is

$\hat{v}={\displaystyle \frac{dr}{dt}}={\displaystyle \frac{d}{dt}}(a[\cos \omega t\hat{i}+\sin \omega t\hat{j}+\omega t\hat{k}])$

$\overrightarrow{v}=-a\omega \sin \omega t\hat{i}+a\omega \cos \omega t\hat{j}+a\omega \hat{k}$

The magnitude of velocity is

$|v|=\sqrt{v_x^2+v_y^2+v_z^2}$

$|v|=\sqrt{(-a\omega \sin \omega t{)}^2+(a\omega \cos \omega t{)}^2+(a\omega {)}^2}$

$|v|=a\omega \sqrt{(-\sin \omega t{)}^2+(\cos \omega t{)}^2+(1{)}^2}=\sqrt{2}a\omega$

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A particle starts from the origin at t = 0 with a velocity $5.0\hat{i}{\text{ ms}}^{-1}$ and moves in the xy-plane under the action of a force which produces a constant acceleration of $(3.0\hat{i}+2.0\hat{j}){\text{ ms}}^{-2}$. What is the y-coordinate of the particle at the instant its x-coordinate is 84 m?(a) 36 m
(b) 24 m
(c) 39 m
(d) 18 m

Solution.(a)

Given, initial velocity of the particle at t=0 seconds,

$v_0=5.0\hat{i}{\text{ ms}}^{-1}$ and

acceleration, $a=(3.0\hat{i}+2.0\hat{j}){\text{ ms}}^{-2}$.

The position of the particle is given by

$r(t)=v_{0}t+\frac{1}{2}a{t}^2=5.0\hat{i}t+(1/2)(3.0\hat{i}+2.0\hat{j})t^2$

$r(t)=(5.0t+1.5t^2)\hat{i}+1.0t^2\hat{j}\text{….(i)}$

As, $r(t)=x(t)\hat{i}+y(t)\hat{j}\text{….(ii)}$

Comparing Eqs. (i) and (ii), we get

$x(t)=5.0t+1.5t^2\text{and}y(t)=1.0t^2$

Given, x(t) = 84 m

$5.0t+1.5t^2=84\Rightarrow 1.5t^2+5.0t-84=0$

Solving the above quadratic equation, the value of t is given as,

$t={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}$

$t={\displaystyle \frac{-5\pm \sqrt{5^2-4(1.5)(-84)}}{2(1.5)}}$

$t={\displaystyle \frac{-5\pm \sqrt{25+504}}{3}}$

$t={\displaystyle \frac{-5\pm \sqrt{529}}{3}}$

$t={\displaystyle \frac{-5\pm 23}{3}}=6\text{ or }-9.33$

(Neglecting the negative values as time can never be negative)

t = 6 seconds

At $t=6\text{ s}$, $y=1.0(6{)}^2=36\text{ m}$.

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When an object is shot from the bottom of a long smooth inclined plane kept at an angle of 60° with the horizontal, it can travel a distance x₁ along the plane. But when the inclination is decreased to 30° and the same object is shot with the same velocity, it can travel a distance x₂. Then x₁ : x₂ will be:
(a) √2 : 1
(b) 1 : √3
(c) 1 : 2√3
(d) 1 : √2

[NEET 2019]

Solution.(b)

The motion of object shot in two cases can be depicted as below :

Case I: Inclination θ = 60°

Case II: Inclination θ = 30°

Using third equation of motion,

$v^2=u^2-2ah\text{….(i)}$

As the object stops finally, so $v=0$.

For inclined motion, $a=g\sin \theta$ and $h=x$.

Substituting these values in Eq. (i), we get

$u^2=2g\sin \theta \cdot x\Rightarrow x={\displaystyle \frac{u^2}{2g\sin \theta }}$

For Case I, $x_1={\displaystyle \frac{u^2}{2g\sin {60}^{\circ }}}$

For Case II, $x_2={\displaystyle \frac{u^2}{2g\sin {30}^{\circ }}}$

${\displaystyle \frac{x_1}{x_2}}={\displaystyle \frac{u^2}{2g\sin {60}^{\circ }}}\times \frac{2g\sin {30}^{\circ }}{u^2}$

${\displaystyle \frac{x_1}{x_2}}={\displaystyle \frac{\sin {30}^{\circ }}{\sin {60}^{\circ }}}$

${\displaystyle \frac{x_1}{x_2}}={\displaystyle \frac{1/2}{\sqrt{3}/2}}={\displaystyle \frac{1}{\sqrt{3}}}\text{ or }1:\sqrt{3}$

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TOPIC 4: MCQs Based on Topic Relative Velocity in Two Dimensions

If two objects P and Q move along parallel straight lines in opposite directions with velocities $v_P$ and $v_Q$ respectively, then the relative velocity of P w.r.t. Q is:
(a) $v_{PQ}=v_P-v_Q$
(b) $v_P-v_Q$
(c) $v_P+v_Q$
(d) $v_Q-v_P$

Solution.(c)

Relative velocity of P w.r.t. Q is given by

$v_{PQ}=v_P-(-v_Q)=v_P+v_Q$

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Buses A and B are moving in the same direction with velocities $20\hat{i}{\text{ ms}}^{-1}$ and $15\hat{i}{\text{ ms}}^{-1}$, respectively. Then, the relative velocity of A w.r.t. B is:
(a) $35\hat{i}{\text{ ms}}^{-1}$
(b) $5\hat{i}{\text{ ms}}^{-1}$
(c) $5\hat{j}{\text{ ms}}^{-1}$
(d) $35\hat{j}{\text{ ms}}^{-1}$

Solution.(b)

Given, $v_A=20\hat{i}{\text{ ms}}^{-1}$, $v_B=15\hat{i}{\text{ ms}}^{-1}$

Relative velocity of A w.r.t. B:

$v_{AB}=v_A-v_B=20\hat{i}-15\hat{i}=5\hat{i}{\text{ ms}}^{-1}$

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Rain is falling vertically with a speed of 35 m/s. A woman rides a bicycle with a speed of 12 m/s in east to west direction. The direction in which she should hold her umbrella is :
(a) at ${\cos }^{-1}(0.343)$ with vertical towards east
(b) at ${\tan }^{-1}(0.343)$ with vertical towards west
(c) at ${\cos }^{-1}(0.343)$ with vertical towards west
(d) at ${\tan }^{-1}(0.343)$ with vertical towards east

Solution.(b)

Velocity of rain is $v_r$ and $v_b$ is the velocity of the bicycle the woman is riding. Both these velocities are with respect to the ground.

Since the woman is riding a bicycle, the velocity of rain as experienced by her is the velocity of rain relative to the bicycle:

$v_{rb}=v_r-v_b$

The angle $\theta$ made by the relative velocity $v_{rb}$ with the vertical is given by

$\tan \theta ={\displaystyle \frac{v_b}{v_r}}={\displaystyle \frac{12}{35}}=0.343\Rightarrow \theta ={\tan }^{-1}(0.343)$

Therefore, the woman should hold her umbrella at an angle of about ${\tan }^{-1}(0.343)$ with the vertical towards the west.

Frequently linked concepts include Relative Velocity in a Plane: Relative Velocity of Rain w.r.t. Moving Man Solved Examples

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A car driver is moving towards a fired rocket with a velocity of $8\hat{i}{\text{ ms}}^{-1}$. He observed the rocket to be moving with a speed of 10 m/s. A stationary observer will see the rocket to be moving with a speed of :
(a) 5 m/s
(b) 6 m/s
(c) 7 m/s
(d) 8 m/s

Solution.(b)

The velocity of car driver, $v_c=8\hat{i}{\text{ ms}}^{-1}$

Velocity of rocket $=v_y\hat{j}{\text{ ms}}^{-1}$

Relative velocity of rocket w.r.t. car $=8\hat{i}-v_y\hat{j}$

Since the speed of the rocket observed by the car driver is $10{\text{ ms}}^{-1}$:

$(v_y{)}^2+(8{)}^2=(10{)}^2$

$v_y^2=100-64=36\Rightarrow v_y=6{\text{ ms}}^{-1}$

$\text{Velocity of rocket, }v=6\hat{j}{\text{ ms}}^{-1}$

Relative speed of rocket w.r.t. a stationary observer $=6-0=6{\text{ ms}}^{-1}$

Gain deeper understanding by studying JEE Main PYQs Solutions for Vectors (Class 11 Physics Motion in a Plane)

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The stream of a river is flowing with a speed of 2 km/h. A swimmer can swim at a speed of 4 km/h. What should be the direction of the swimmer with respect to the flow of the river to cross the river straight?
(a) 60°
(b) 120°
(c) 90°
(d) 150°

[JEE Main 2019]

Solution.(b)

Let the velocity of the swimmer be v_s = 4 km/h and velocity of river be v_r = 2 km/h.

Also, angle of swimmer with the flow of the river (downstream) is α = 90°+θ.

From diagram, angle θ is:

$\sin \theta ={\displaystyle \frac{v_r}{v_s}}$

$\sin \theta ={\displaystyle \frac{2\text{ km/h}}{4\text{ km/h}}}$

$\sin \theta ={\displaystyle \frac{1}{2}}\Rightarrow \theta ={30}^{\circ }$

$\alpha ={90}^{\circ }+{30}^{\circ }={120}^{\circ }$

For complete preparation, also study JEE Main Vectors Chapterwise PYQs Solutions (Motion in a Plane)

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A man standing on a road has to hold his umbrella at 30° with the vertical to keep the rain away. He throws the umbrella and starts running at 10 km/h. He finds that raindrops are hitting his head vertically. The actual speed of the raindrops is:
(a) 20 km/h
(b) 10√3 km/h
(c) 20√3 km/h
(d) 10 km/h

Solution.(a)

When the man is at rest with respect to the ground, the rain comes to him at an angle 30° with the vertical.

Here, $v_{rg}=$ velocity of the rain with respect to the ground,

$v_{mg}=$ velocity of the man with respect to the ground.

When the man throws the umbrella and starts running, then

$|v_{mg}|=10{\text{ kmh}}^{-1}$.

From vector components:

$\sin {30}^{\circ }={\displaystyle \frac{v_{mg}}{v_{rg}}}\Rightarrow v_{rg}={\displaystyle \frac{10}{\sin {30}^{\circ }}}=20{\text{ kmh}}^{-1}$

Practice more questions from NEET PYQs Solutions for Vectors (Class 11 Physics Motion in a Plane)

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A girl can swim with a speed of 5 km/h in still water. She crosses a river 2 km wide, where the river flows steadily at 2 km/h and she makes strokes normal to the river current. Find how far down the river she goes when she reaches the other bank.
(a) 1 km
(b) 2 km
(c) 800 m
(d) 750 m

Solution.(c)

Given, speed of girl, $v_g=5{\text{ km h}}^{-1}$

Speed of river, $v_r=2{\text{ km h}}^{-1}$

Width of river, $d=2\text{ km}$

Since the girl crosses the river normal to the flow of the river, time taken by the girl to cross the river:

$t={\displaystyle \frac{d}{v_g}}$

$t={\displaystyle \frac{2\text{ km}}{5{\text{ kmh}}^{-1}}}$

$t={\displaystyle \frac{2}{5}}\text{ h}$

In this time, the girl will go down the river by the distance AC due to river current.

$\text{Distance travelled along the river }=v_r\times t=2\times {\displaystyle \frac{2}{5}}$

$={\displaystyle \frac{4}{5}}\text{ km}=800\text{ m}$

Enhance your preparation with NCERT Solutions for Vectors Class 11 Physics Chapter Motion in a Plane

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A girl riding a bicycle with a speed of 5 m/s towards east direction sees raindrops falling vertically downwards. On increasing her speed to 15 m/s, rain appears to fall making an angle of 45° with the vertical. Find the magnitude of the velocity of the rain.
(a) 5 m/s
(b) 5√5 m/s
(c) 25 m/s
(d) 10 m/s

Solution.(b)

Given, velocity of girl, $v_g=5\hat{i}{\text{ ms}}^{-1}$

Let velocity of rain, $v_r=v_x\hat{i}+v_y\hat{j}{\text{ ms}}^{-1}$

Relative velocity of rain $=v_r-v_g=(v_x-5)\hat{i}+v_y\hat{j}$

Now, it is vertical, so

$\tan \theta ={\displaystyle \frac{v_x-5}{v_y}}=0\Rightarrow v_x-5=0\Rightarrow v_x=5\text{….(i)}$

On increasing the speed of the girl to $15{\text{ ms}}^{-1}$, relative velocity becomes $(v_x-15)\hat{i}+v_y\hat{j}$.

$\tan {45}^{\circ }={\displaystyle \frac{v_x-15}{v_y}}=1$

$v_x-15=v_y$

$5-15=v_y$

$v_y=-10\text{[using Eq. (i)]}$

$\text{Velocity of rain }=(5\hat{i}-10\hat{j}){\text{ ms}}^{-1}$

$\text{Magnitude of velocity of rain }=\sqrt{(5{)}^2+(10{)}^2}$

$=\sqrt{125}=5\sqrt{5}{\text{ ms}}^{-1}$

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