Q: A force of 10 N acts at 60^\circ to the direction of a 4 m displacement. The work done is

20 J. W = Fd\cos\theta = 10\times4\times\cos 60^\circ = 20 J.

Q: A particle rotates with angular velocity \omega = 2 rad/s at position r = 3 m perpendicular to the rotation axis. Its speed |\vec v| = |\vec\omega\times\vec r| is

6 m/s. v = \omega r\sin 90^\circ = 2\times3 = 6 m/s.

Q: For unit vectors \hat{p} and \hat{q}, \hat{p}\cdot\hat{q} equals

\cos\theta, where \theta is the angle between them. \hat{p}\cdot\hat{q} = (1)(1)\cos\theta = \cos\theta.

Question 1

The value of \lambda for which \vec A = \lambda\hat{i} + \hat{j} is perpendicular to \vec B = \hat{i} - \hat{j} is

Accepted Answer

1. \vec A\cdot\vec B = \lambda - 1 = 0 \Rightarrow \lambda = 1.

Question 2

A body of mass 0.5 kg has momentum \vec p = (3\hat{i} + 4\hat{j}) kg·m/s. Its kinetic energy is

Accepted Answer

25 J. p^2 = \vec p\cdot\vec p = 9+16 = 25; KE = \frac{p^2}{2m} = \frac{25}{1} = 25 J.

Question 3

Forces \vec F_1 = \hat{i}+\hat{j} N and \vec F_2 = 2\hat{i}-\hat{j} N act on a body displaced by \vec d = 3\hat{i} m. The total work done is

Accepted Answer

9 J. Net force = 3\hat{i} N; W = (3\hat{i})\cdot(3\hat{i}) = 9 J.

Question 4

\vec B is parallel to \vec A = 4\hat{i} + 3\hat{j} and |\vec B| = 10. Then \vec B is

Accepted Answer

8\hat{i} + 6\hat{j}. \hat{A} = \frac{4\hat{i}+3\hat{j}}{5}; \vec B = 10\hat{A} = 8\hat{i}+6\hat{j}.

Question 5

A force of 10 N acts at 60^\circ to the direction of a 4 m displacement. The work done is

Accepted Answer

20 J. W = Fd\cos	heta = 10	imes4	imes\cos 60^\circ = 20 J.

Question 6

A particle rotates with angular velocity \omega = 2 rad/s at position r = 3 m perpendicular to the rotation axis. Its speed |\vec v| = |\vec\omega	imes\vec r| is

Accepted Answer

6 m/s. v = \omega r\sin 90^\circ = 2	imes3 = 6 m/s.

Question 7

What vector must be added to \hat{i} + 2\hat{j} to get a resultant of 3\hat{i}?

Accepted Answer

2\hat{i} - 2\hat{j}. Required vector = 3\hat{i} - (\hat{i}+2\hat{j}) = 2\hat{i} - 2\hat{j}.

Question 8

The vectors \vec A = \hat{i} + \hat{j} - 2\hat{k} and \vec B = 2\hat{i} - 2\hat{j} are

Accepted Answer

Perpendicular. \vec A\cdot\vec B = (1)(2) + (1)(-2) + (-2)(0) = 0 — perpendicular.

Question 9

For unit vectors \hat{p} and \hat{q}, \hat{p}\cdot\hat{q} equals

Accepted Answer

\cos	heta, where 	heta is the angle between them. \hat{p}\cdot\hat{q} = (1)(1)\cos	heta = \cos	heta.

Question 10

If (\vec a + \vec b)\cdot(\vec a + \vec b) = a^2 + b^2, then \vec a and \vec b are

Accepted Answer

Perpendicular. Expanding gives a^2 + b^2 + 2\vec a\cdot\vec b; equality forces \vec a\cdot\vec b = 0.

Question 11

For a line making angles \alpha, \beta, \gamma with the coordinate axes, \sin^2\alpha + \sin^2\beta + \sin^2\gamma equals

Accepted Answer

2. \sum\sin^2 = \sum(1 - \cos^2) = 3 - 1 = 2.

Question 12

Can a line make angles 30^\circ, 45^\circ and 60^\circ with the x, y and z axes respectively?

Accepted Answer

No, because \cos^2 30^\circ + \cos^2 45^\circ + \cos^2 60^\circ 
eq 1. \frac{3}{4} + \frac{1}{2} + \frac{1}{4} = \frac{3}{2} 
eq 1 — such a line cannot exist.

Question 13

If a vector makes \gamma = 90^\circ with the z-axis, the vector lies

Accepted Answer

In the xy-plane. n = \cos 90^\circ = 0 means no z-component — the vector lies entirely in the xy-plane.

Question 14

Two equal vectors of magnitude 10 units act at 120° to each other. Their resultant is:

Accepted Answer

10 units. R = √(A²+A²+2A²cos120°) = √(100+100−100) = 10 units.

Question 15

Two vectors of 10 units each act at 60°. Their resultant is:

Accepted Answer

10√3 units. R = √(100+100+2(100)(0.5)) = √300 = 10√3 units.

Question 16

Two equal vectors of magnitude 6 act at 60°. The magnitude of their difference is:

Accepted Answer

6. |A−B| = √(A²+A²−2A²cos60°) = √(36+36−36) = 6.

Question 17

Two vectors of magnitude 4 and 5 have a dot product of 10. The angle between them is:

Accepted Answer

60°. cosθ = 10/(4×5) = 0.5 ⇒ θ = 60°.

Question 18

The area of a triangle with two sides 10 and 10 at 60° is:

Accepted Answer

25√3. Area = ½ab sinθ = ½(10)(10)(√3/2) = 25√3.

Question 19

The direction cosines of the vector 3î + 4k̂ are:

Accepted Answer

(3/5, 0, 4/5). |A| = 5; l = 3/5, m = 0, n = 4/5.

Question 20

A vector makes equal angles with all three axes. Each direction cosine is:

Accepted Answer

1/√3. 3l² = 1 ⇒ l = 1/√3.

Vectors — JEE Main Physics MCQs with Solutions

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Sample questions with solutions