Q: A cyclist rides along a circular track at constant speed v. After turning through 60^\circ of arc, the magnitude of the change in velocity is

v. |\Delta\vec v| = 2v\sin(\theta/2) = 2v\sin 30^\circ = v.

Q: \hat{i}\cdot(\hat{j}\times\hat{k}) equals

1. \hat{j}\times\hat{k} = \hat{i}, and \hat{i}\cdot\hat{i} = 1.

Q: If |\vec A| = 4, |\vec B| = 3 and the angle between them is 30^\circ, then |\vec A \times \vec B| is

6. |\vec A\times\vec B| = AB\sin\theta = 4\times 3\times\tfrac{1}{2} = 6.

Q: If |\vec A| = 4, |\vec B| = 3 and the angle between them is 60^\circ, then \vec A \cdot \vec B is

6. \vec A\cdot\vec B = AB\cos\theta = 4\times 3\times\tfrac{1}{2} = 6.

Q: If \vec A \cdot \vec B = |\vec A \times \vec B|, the angle between \vec A and \vec B is

45^\circ. AB\cos\theta = AB\sin\theta \Rightarrow \tan\theta = 1 \Rightarrow \theta = 45^\circ.

Question 1

Two vectors each of magnitude 5 units are inclined at 60^\circ. The magnitude of their difference is

Accepted Answer

5. |\vec A - \vec B| = 2(5)\sin(60^\circ/2) = 10\sin 30^\circ = 5 units.

Question 2

Two vectors have magnitudes 7 and 4 units. Which of the following CANNOT be the magnitude of \vec A - \vec B?

Accepted Answer

2. The difference must lie between 3 and 11; a value of 2 is impossible.

Question 3

A cyclist rides along a circular track at constant speed v. After turning through 60^\circ of arc, the magnitude of the change in velocity is

Accepted Answer

v. |\Delta\vec v| = 2v\sin(	heta/2) = 2v\sin 30^\circ = v.

Question 4

If \vec A = 2\hat{i} + 3\hat{j} and \vec B = \hat{i} - \hat{j}, then \vec A \cdot \vec B is

Accepted Answer

-1. \vec A\cdot\vec B = (2)(1) + (3)(-1) = 2 - 3 = -1.

Question 5

A force \vec F = (3\hat{i} + 4\hat{j}) N displaces a body by \vec d = (2\hat{i} + \hat{j}) m. The work done is

Accepted Answer

10 J. W = \vec F\cdot\vec d = (3)(2) + (4)(1) = 10 J.

Question 6

\hat{i}\cdot(\hat{j}	imes\hat{k}) equals

Accepted Answer

1. \hat{j}	imes\hat{k} = \hat{i}, and \hat{i}\cdot\hat{i} = 1.

Question 7

If |\vec A| = 4, |\vec B| = 3 and the angle between them is 30^\circ, then |\vec A 	imes \vec B| is

Accepted Answer

6. |\vec A	imes\vec B| = AB\sin	heta = 4	imes 3	imes	frac{1}{2} = 6.

Question 8

If |\vec A| = 4, |\vec B| = 3 and the angle between them is 60^\circ, then \vec A \cdot \vec B is

Accepted Answer

6. \vec A\cdot\vec B = AB\cos	heta = 4	imes 3	imes	frac{1}{2} = 6.

Question 9

If \vec A \cdot \vec B = |\vec A 	imes \vec B|, the angle between \vec A and \vec B is

Accepted Answer

45^\circ. AB\cos	heta = AB\sin	heta \Rightarrow 	an	heta = 1 \Rightarrow 	heta = 45^\circ.

Question 10

If |\vec A| = 5 and |\vec B| = 3, then (\vec A + \vec B)\cdot(\vec A - \vec B) equals

Accepted Answer

16. (\vec A+\vec B)\cdot(\vec A-\vec B) = A^2 - B^2 = 25 - 9 = 16.

Question 11

(\vec A 	imes \vec B)^2 + (\vec A \cdot \vec B)^2 equals

Accepted Answer

A^2B^2. A^2B^2\sin^2	heta + A^2B^2\cos^2	heta = A^2B^2(\sin^2	heta + \cos^2	heta) = A^2B^2.

Question 12

A unit vector perpendicular to both \hat{i} and \hat{j} is

Accepted Answer

\hat{k}. \hat{i}	imes\hat{j} = \hat{k}, which is perpendicular to both (-\hat{k} also works).

Question 13

A force \vec F = (2\hat{i} + 3\hat{j}) N acts on a body moving with velocity \vec v = (\hat{i} + 2\hat{j}) m/s. The power delivered is

Accepted Answer

8 W. P = \vec F\cdot\vec v = (2)(1) + (3)(2) = 8 W.

Question 14

If \vec A \cdot \vec B = -|\vec A||\vec B|, the angle between the vectors is

Accepted Answer

180^\circ. \cos	heta = -1 \Rightarrow 	heta = 180^\circ (antiparallel vectors).

Question 15

If \hat{a} and \hat{b} are unit vectors with \hat{a}\cdot\hat{b} = 1, then

Accepted Answer

They point in the same direction. \hat{a}\cdot\hat{b} = \cos	heta = 1 \Rightarrow 	heta = 0^\circ: identical directions.

Question 16

The area of a parallelogram with sides of lengths 5 and 4 units inclined at 30^\circ is

Accepted Answer

10 sq units. Area = ab\sin	heta = 5	imes4	imes\sin 30^\circ = 10 sq units.

Question 17

The area of the triangle whose two sides are the vectors 3\hat{i} and 4\hat{j} is

Accepted Answer

6 sq units. Area = 	frac{1}{2}|\vec A	imes\vec B| = 	frac{1}{2}(3)(4)\sin 90^\circ = 6 sq units.

Question 18

The diagonals of a parallelogram are 2\hat{i} and 4\hat{j}. Its area is

Accepted Answer

4 sq units. Area = 	frac{1}{2}|\vec d_1	imes\vec d_2| = 	frac{1}{2}(2)(4) = 4 sq units.

Question 19

A parallelogram has sides 3 and 4 units at right angles. Its area is

Accepted Answer

12 sq units. Area = ab\sin 90^\circ = 3	imes4 = 12 sq units (a rectangle).

Question 20

Two sides of a triangle are equal vectors of magnitude a with angle 	heta between them. The area of the triangle is

Accepted Answer

\frac{1}{2}a^2\sin	heta. Area = 	frac{1}{2}|\vec A	imes\vec B| = 	frac{1}{2}a\cdot a\sin	heta.

Vectors — JEE Main Physics MCQs with Solutions

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Sample questions with solutions