Vectors — JEE Main Physics MCQs with Solutions
Free JEE Main Physics Vectors MCQs with step-by-step solutions (13 questions). Part of Motion in a Plane. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Vectors · medium · numerical
If the magnitude of sum of two vectors is equal to the magnitude of difference of the two vectors, the angle between these vectors is
A. $90^\circ$ ✓ Correct
B. $45^\circ$
C. $180^\circ$
D. $0^\circ$
Solution: Let the two vectors be $P$ and $Q$. Given $|P+Q|=|P-Q|$ with angle $\phi$ between them. Then $P^2+Q^2+2PQ\cos\phi=P^2+Q^2-2PQ\cos\phi\Rightarrow 4PQ\cos\phi=0\Rightarrow\cos\phi=0\Rightarrow\phi=\dfrac{\pi}{2}=90^\circ$.
Q2 — Vectors · medium · numerical
If vectors $A=\cos\omega t\,\hat{i}+\sin\omega t\,\hat{j}$ and $B=\cos\dfrac{\omega t}{2}\,\hat{i}+\sin\dfrac{\omega t}{2}\,\hat{j}$ are functions of time, then the value of $t$ at which they are orthogonal to each other, is
A. $t=\dfrac{\pi}{4\omega}$
B. $t=\dfrac{\pi}{2\omega}$
C. $t=\dfrac{\pi}{\omega}$ ✓ Correct
D. $t=0$
Solution: For perpendicular vectors, $A\cdot B=0$. So $[\cos\omega t\,\hat{i}+\sin\omega t\,\hat{j}]\cdot[\cos\dfrac{\omega t}{2}\,\hat{i}+\sin\dfrac{\omega t}{2}\,\hat{j}]=0\Rightarrow\cos\omega t\cos\dfrac{\omega t}{2}+\sin\omega t\sin\dfrac{\omega t}{2}=0$. Using $\cos(A-B)=\cos A\cos B+\sin A\sin B$, $\cos\left(\omega t-\dfrac{\omega t}{2}\right)=0\Rightarrow\cos\dfrac{\omega t}{2}=0\Rightarrow\dfrac{\omega t}{2}=\dfrac{\pi}{2}\Rightarrow t=\dfrac{\pi}{\omega}$.
Q3 — Vectors · medium · theory
Six vectors $a$ to $f$ have the magnitudes and directions indicated in the figure. Which of the following statements is true?
A. $b+c=f$
B. $d+c=f$
C. $d+e=f$ ✓ Correct
D. $b+e=f$
Solution: If two non-zero vectors are represented by the two adjacent sides of a parallelogram, the resultant is given by the diagonal passing through their point of intersection. Hence $d+e=f$.
Q4 — Vectors · medium · numerical
$A$ and $B$ are two vectors and $\theta$ is the angle between them. If $|A\times B|=\sqrt{3}(A\cdot B)$, then the value of $\theta$ is
A. $60^\circ$ ✓ Correct
B. $45^\circ$
C. $30^\circ$
D. $90^\circ$
Solution: Given $|A\times B|=\sqrt{3}(A\cdot B)\Rightarrow AB\sin\theta=\sqrt{3}\,AB\cos\theta\Rightarrow\tan\theta=\sqrt{3}\Rightarrow\theta=60^\circ$.
Q5 — Vectors · medium · numerical
If a vector $2\hat{i}+3\hat{j}+8\hat{k}$ is perpendicular to the vector $4\hat{j}-4\hat{i}+\alpha\hat{k}$, then the value of $\alpha$ is
A. $-1$
B. $\dfrac{1}{2}$
C. $-\dfrac{1}{2}$ ✓ Correct
D. $1$
Solution: If two vectors are perpendicular, their dot product is zero. Let $a=2\hat{i}+3\hat{j}+8\hat{k}$ and $b=-4\hat{i}+4\hat{j}+\alpha\hat{k}$. Then $a\cdot b=0\Rightarrow(2\hat{i}+3\hat{j}+8\hat{k})\cdot(-4\hat{i}+4\hat{j}+\alpha\hat{k})=0\Rightarrow-8+12+8\alpha=0\Rightarrow 8\alpha=-4\Rightarrow\alpha=-\dfrac{4}{8}=-\dfrac{1}{2}$.
Q6 — Vectors · medium · numerical
If $|A\times B|=\sqrt{3}\,A\cdot B$, then the value of $|A+B|$ is
A. $(A^2+B^2+AB)^{1/2}$ ✓ Correct
B. $\left(A^2+B^2+\dfrac{AB}{\sqrt{3}}\right)^{1/2}$
C. $A+B$
D. $(A^2+B^2+\sqrt{3}AB)^{1/2}$
Solution: Given $|A\times B|=\sqrt{3}\,A\cdot B$. Since $|A\times B|=AB\sin\theta$ and $A\cdot B=AB\cos\theta$, we get $AB\sin\theta=\sqrt{3}\,AB\cos\theta\Rightarrow\tan\theta=\sqrt{3}\Rightarrow\theta=60^\circ$. By the parallelogram law, $|A+B|=\sqrt{A^2+B^2+2AB\cos60^\circ}=\sqrt{A^2+B^2+2AB\times\dfrac{1}{2}}=(A^2+B^2+AB)^{1/2}$.
Q7 — Vectors · medium · theory
The vector sum of two forces is perpendicular to their vector differences. In that case, the forces
A. are not equal to each other in magnitude
B. cannot be predicted
C. are equal to each other
D. are equal to each other in magnitude ✓ Correct
Solution: Let $A$ and $B$ be the two forces. Sum $F_1=A+B$ and difference $F_2=A-B$. Since the sum is perpendicular to the difference, $F_1\cdot F_2=0\Rightarrow(A+B)\cdot(A-B)=0\Rightarrow A^2-B^2=0\Rightarrow A^2=B^2\Rightarrow|A|=|B|$. Thus the forces are equal in magnitude.
Q8 — Vectors · medium · numerical
If a unit vector is represented by $0.5\hat{i}+0.8\hat{j}+c\hat{k}$, then the value of $c$ is
A. $1$
B. $\sqrt{0.11}$ ✓ Correct
C. $\sqrt{0.01}$
D. $0.39$
Solution: For a unit vector, $|\hat{n}|=1$. So $\sqrt{(0.5)^2+(0.8)^2+c^2}=1\Rightarrow 0.25+0.64+c^2=1\Rightarrow 0.89+c^2=1\Rightarrow c^2=1-0.89=0.11\Rightarrow c=\sqrt{0.11}$.
Q9 — Vectors · medium · theory
Which of the following is not a vector quantity?
A. Speed ✓ Correct
B. Velocity
C. Torque
D. Displacement
Solution: Speed is a scalar quantity as it gives no information about the direction of motion. Velocity, displacement and torque each possess both magnitude and direction, so they are vector quantities. Hence speed is not a vector quantity.
Q10 — Vectors · medium · numerical
The angle between the two vectors $A=3\hat{i}+4\hat{j}+5\hat{k}$ and $B=3\hat{i}+4\hat{j}-5\hat{k}$ will be
A. $0^\circ$
B. $45^\circ$
C. $90^\circ$ ✓ Correct
D. $180^\circ$
Solution: $\cos\theta=\dfrac{A\cdot B}{AB}$. Here $A=\sqrt{3^2+4^2+5^2}=\sqrt{50}$, $B=\sqrt{3^2+4^2+(-5)^2}=\sqrt{50}$ and $A\cdot B=9+16-25=0$. So $\cos\theta=\dfrac{0}{\sqrt{50}\cdot\sqrt{50}}=0\Rightarrow\theta=90^\circ$.
Q11 — Vectors · medium · theory
The resultant of $A\times 0$ will be equal to
A. zero
B. $A$
C. zero vector ✓ Correct
D. unit vector
Solution: From the properties of the vector product, the cross product of any vector with the zero vector is a null (zero) vector.
Q12 — Vectors · medium · theory
The angle between $A$ and $B$ is $\theta$. The value of the triple product $A\cdot(B\times A)$ is
A. $A^2B$
B. zero ✓ Correct
C. $A^2B\sin\theta$
D. $A^2B\cos\theta$
Solution: In a scalar triple product the positions of dot and cross can be interchanged: $A\cdot(B\times A)=(A\times B)\cdot A=(A\times A)\cdot B$. But $A\times A=0$, so $A\cdot(B\times A)=0$. Alternatively, if $A\times B=C$ then $C$ is perpendicular to $A$, so $A\cdot C=0$.
Q13 — Vectors · medium · numerical
The magnitudes of vectors $\vec{A}$, $\vec{B}$ and $\vec{C}$ are 3, 4 and 5 units respectively. If $\vec{A}+\vec{B}=\vec{C}$, the angle between $\vec{A}$ and $\vec{B}$ is
A. $\dfrac{\pi}{2}$ ✓ Correct
B. $\cos^{-1}(0.6)$
C. $\tan^{-1}\left(\dfrac{7}{5}\right)$
D. $\dfrac{\pi}{4}$
Solution: Given $|\vec{A}|=3$, $|\vec{B}|=4$, $|\vec{C}|=5$ and $\vec{A}+\vec{B}=\vec{C}$. So $5^2=3^2+4^2+2\cdot4\cdot3\cos\theta \Rightarrow \cos\theta=0 \Rightarrow \theta=\dfrac{\pi}{2}$. Thus $\vec{A}$ is perpendicular to $\vec{B}$.