Vector (Cross) Product — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Vector (Cross) Product MCQs with step-by-step solutions (21 questions). Part of Mathematical Methods (Std 11). Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Vector (Cross) Product · easy · theory
The vector product of two vectors is defined as:
A. $AB\cos\theta$, a scalar
B. $AB\sin\theta\,\hat{n}$, a vector perpendicular to both ✓ Correct
C. $\dfrac{A}{B}\sin\theta$
D. $AB\tan\theta\,\hat{n}$
Solution: The unit vector $\hat{n}$ is given by the right-hand rule.
Q2 — Vector (Cross) Product · easy · theory
The result of a vector product of two vectors is:
A. A scalar
B. A vector in the plane containing them
C. A vector perpendicular to the plane containing them ✓ Correct
D. Always a null vector
Solution: This is why it is also called the cross product.
Q3 — Vector (Cross) Product · medium · theory
The vector product is anti-commutative, meaning that:
A. $\vec{A} \times \vec{B} = \vec{B} \times \vec{A}$
B. $\vec{A} \times \vec{B} = -\vec{B} \times \vec{A}$ ✓ Correct
C. $\vec{A} \times \vec{B} = \vec{A}\cdot\vec{B}$
D. $\vec{A} \times \vec{B} = 0$
Solution: Reversing the order reverses the direction of the resulting vector.
Q4 — Vector (Cross) Product · easy · theory
Two non-zero vectors are parallel if their vector product is:
A. Equal to $AB$
B. A null vector ✓ Correct
C. Perpendicular to both
D. Maximum
Solution: With $\theta = 0^\circ$ or $180^\circ$, $\sin\theta = 0$.
Q5 — Vector (Cross) Product · easy · theory
Torque is an example of a:
A. Scalar product of force and velocity
B. Vector product of position vector and force ✓ Correct
C. Vector sum of forces
D. Scalar product of position vector and force
Solution: $\vec{\tau} = \vec{r} \times \vec{F}$.
Q6 — Vector (Cross) Product · medium · theory
Angular momentum is expressed as a vector product in the form:
A. $\vec{L} = \vec{r} \times \vec{p}$ ✓ Correct
B. $\vec{L} = \vec{p} \times \vec{r}$
C. $\vec{L} = \vec{r} + \vec{p}$
D. $\vec{L} = \vec{r}\cdot\vec{p}$
Solution: It is perpendicular to both the position vector and the momentum.
Q7 — Vector (Cross) Product · hard · theory
The magnitude of the vector product $|\vec{A} \times \vec{B}|$ represents geometrically the:
A. Length of the diagonal
B. Area of the parallelogram with $\vec{A}$ and $\vec{B}$ as adjacent sides ✓ Correct
C. Volume of a cube of side $A$
D. Perimeter of that parallelogram
Solution: Half this value gives the area of the triangle formed by the two vectors.
Q8 — Vector (Cross) Product · medium · theory
The direction of $\vec{A} \times \vec{B}$ is given by:
A. Fleming's left-hand rule
B. The direction of $\vec{B}$
C. The right-hand rule applied from $\vec{A}$ to $\vec{B}$ ✓ Correct
D. The direction of $\vec{A}$
Solution: Curling the fingers from the first vector to the second makes the thumb point along the product.
Q9 — Vector (Cross) Product · medium · theory
The vector product of two equal vectors is:
A. A null vector ✓ Correct
B. Equal to $A^2$
C. Equal to $2A$
D. Perpendicular to both
Solution: The angle between them is zero, so $\sin\theta = 0$.
Q10 — Vector (Cross) Product · easy · numerical
Two vectors of magnitudes $3$ and $4$ act at $90^\circ$. The magnitude of their vector product is:
A. Zero
B. $7$
C. $12$ ✓ Correct
D. $5$
Solution: $|\vec{A} \times \vec{B}| = AB\sin 90^\circ = 12$.
Q11 — Vector (Cross) Product · medium · numerical
Two vectors of magnitudes $3$ and $4$ act at $30^\circ$. The magnitude of their vector product is:
A. $5$
B. $6$ ✓ Correct
C. $10.4$
D. $12$
Solution: $3 \times 4 \times \sin 30^\circ = 12 \times 0.5 = 6$.
Q12 — Vector (Cross) Product · medium · numerical
If the vector product of two non-zero vectors is a null vector, the vectors are:
A. Parallel or antiparallel ✓ Correct
B. Perpendicular
C. Inclined at $45^\circ$
D. Of equal magnitude
Solution: $\sin\theta = 0$ requires $\theta = 0^\circ$ or $180^\circ$.
Q13 — Vector (Cross) Product · medium · numerical
The value of $\hat{i} \times \hat{j}$ is:
A. $1$
B. $-\hat{k}$
C. $\hat{k}$ ✓ Correct
D. $0$
Solution: The right-hand rule takes $x$ crossed into $y$ to give $z$.
Q14 — Vector (Cross) Product · medium · numerical
The value of $\hat{j} \times \hat{i}$ is:
A. $1$
B. $\hat{k}$
C. $0$
D. $-\hat{k}$ ✓ Correct
Solution: Reversing the order reverses the direction of the product.
Q15 — Vector (Cross) Product · easy · numerical
The value of $\hat{i} \times \hat{i}$ is:
A. $1$
B. $\hat{k}$
C. A null vector ✓ Correct
D. $\hat{i}$
Solution: A vector crossed with itself gives zero because the angle between them is zero.
Q16 — Vector (Cross) Product · hard · numerical
A force $\vec{F} = 3\hat{j}\text{ N}$ acts at a position $\vec{r} = 2\hat{i}\text{ m}$. The torque about the origin is:
A. $6\hat{k}\text{ N}\cdot\text{m}$ ✓ Correct
B. $6\text{ N}\cdot\text{m}$, a scalar
C. $6\hat{i}\text{ N}\cdot\text{m}$
D. $-6\hat{k}\text{ N}\cdot\text{m}$
Solution: $\vec{\tau} = \vec{r} \times \vec{F} = 2\hat{i} \times 3\hat{j} = 6(\hat{i} \times \hat{j}) = 6\hat{k}$.
Q17 — Vector (Cross) Product · hard · numerical
The cross product of $2\hat{i}$ and $3\hat{k}$ is:
A. $-6\hat{j}$ ✓ Correct
B. $6\hat{j}$
C. A null vector
D. $6\hat{k}$
Solution: $\hat{i} \times \hat{k} = -\hat{j}$, so the result is $-6\hat{j}$.
Q18 — Vector (Cross) Product · hard · numerical
For two vectors, $|\vec{A} \times \vec{B}| = \vec{A}\cdot\vec{B}$. The angle between them is:
A. $60^\circ$
B. $90^\circ$
C. $45^\circ$ ✓ Correct
D. $30^\circ$
Solution: The condition gives $AB\sin\theta = AB\cos\theta$, so $\tan\theta = 1$.
Q19 — Vector (Cross) Product · hard · numerical
For any two vectors, the quantity $|\vec{A} \times \vec{B}|^2 + (\vec{A}\cdot\vec{B})^2$ equals:
A. $A^2B^2$ ✓ Correct
B. $A^2 + B^2$
C. Zero
D. $2A^2B^2$
Solution: It is $A^2B^2(\sin^2\theta + \cos^2\theta) = A^2B^2$.
Q20 — Vector (Cross) Product · medium · numerical
A force of $10\text{ N}$ acts perpendicular to a position vector of length $0.5\text{ m}$. The magnitude of the torque is:
A. $20\text{ N}\cdot\text{m}$
B. $10\text{ N}\cdot\text{m}$
C. $5\text{ N}\cdot\text{m}$ ✓ Correct
D. $0.05\text{ N}\cdot\text{m}$
Solution: $\tau = rF\sin 90^\circ = 0.5 \times 10 = 5\text{ N}\cdot\text{m}$.
Q21 — Vector (Cross) Product · hard · numerical
The area of the triangle formed by two vectors $\vec{A}$ and $\vec{B}$ as adjacent sides is:
A. $|\vec{A} \times \vec{B}|$
B. $\dfrac{1}{2}\vec{A}\cdot\vec{B}$
C. $2|\vec{A} \times \vec{B}|$
D. $\dfrac{1}{2}|\vec{A} \times \vec{B}|$ ✓ Correct
Solution: The cross product gives the parallelogram area, and the triangle is half of it.