CBSE Class 11 Physics Vectors Notes Set 02

Practice Questions

Question 1. Vectors \( \vec{A}, \vec{B} \) and \( \vec{C} \) are shown in the figure. Find the angle between (i) \( \vec{A} \) and \( \vec{B} \), (ii) \( \vec{A} \) and \( \vec{C} \), (iii) \( \vec{B} \) and \( \vec{C} \).

Answer:
Measured anticlockwise from the +x axis, \( \vec{A} \) is at 30°, \( \vec{B} \) at 90° + 45° = 135° and \( \vec{C} \) at 180° + 60° = 240°.
(i) Angle between \( \vec{A} \) and \( \vec{B} \) = 135° − 30° = 105°
(ii) Angle between \( \vec{A} \) and \( \vec{C} \) = 240° − 30° = 210°, so the smaller angle is 360° − 210° = 150°
(iii) Angle between \( \vec{B} \) and \( \vec{C} \) = 240° − 135° = 105°

Question 2. Two forces, each numerically equal to 5 N, are acting as shown in the figure. Find the angle between the forces.

Answer:
The inclined force makes 60° with the positive x-direction, while the horizontal force acts along the negative x-direction (towards the left).
Placing them tail to tail, the angle between the forces = 180° − 60° = 120°.

Question 3. Rain is falling vertically downwards with a speed of 5 m/s. If the unit vector along upward is defined as \( \hat{j} \), represent the velocity of rain in vector form.
Answer:
Velocity is 5 m/s opposite to \( \hat{j} \), so \( \vec{v}_R = -5\hat{j} \) m/s.

Question 4. A man walks 40 m north, then 30 m east and then 40 m south. Find the displacement from the starting point.
Answer:
Net northward displacement = 40 − 40 = 0; eastward displacement = 30 m.
Displacement = 30 m towards east.

Question 5. Two forces \( \vec{F}_1 \) and \( \vec{F}_2 \) are acting at right angles to each other. Find their resultant.
Answer:
With θ = 90°, \( R = \sqrt{F_1^2 + F_2^2 + 2F_1F_2\cos 90^\circ} = \sqrt{F_1^2 + F_2^2} \)

Question 6. A vector of magnitude 30 and direction eastwards is added to another vector of magnitude 40 and direction northwards. Find the magnitude and the direction of the resultant with the east.
Answer:
\( R = \sqrt{30^2 + 40^2} = 50 \)
\( \tan\theta = \frac{40}{30} = \frac{4}{3} \) ⇒ θ = 53° with the east (north of east).

Question 7. Two forces \( \vec{F}_1 \) = 500 N due east and \( \vec{F}_2 \) = 250 N due north. Find \( \vec{F}_2 - \vec{F}_1 \).
Answer:
\( \vec{F}_2 - \vec{F}_1 \) = 250 N north + 500 N west (the two parts are perpendicular).
Magnitude \( = \sqrt{250^2 + 500^2} = 250\sqrt{5} \) N
Direction: \( \tan\theta = \frac{500}{250} = 2 \), i.e. at tan⁻¹(2) west of north.

Question 8. Find the magnitude of \( 3\hat{i} + 2\hat{j} + \hat{k} \).
Answer:
\( \sqrt{3^2 + 2^2 + 1^2} = \sqrt{14} \)

Question 9. If \( \vec{A} = 3\hat{i} + 4\hat{j} \), then find \( \hat{A} \).
Answer:
\( |\vec{A}| = \sqrt{9 + 16} = 5 \), so \( \hat{A} = \frac{\vec{A}}{|\vec{A}|} = \frac{3\hat{i} + 4\hat{j}}{5} \)

Question 10. What are the x and the y components of a 25 m displacement at an angle of 210° with the x-axis (clockwise)?
Answer:
210° clockwise from the +x axis is the same as 150° anticlockwise.
x-component = 25 cos 150° = −25 cos 30°; y-component = 25 sin 150° = +25 sin 30°
i.e. x = −25 cos 30° ≈ −21.65 m and y = +25 sin 30° = 12.5 m

Question 11. One of the rectangular components of a velocity of 60 km h⁻¹ is 30 km h⁻¹. Find the other rectangular component.
Answer:
\( v_2 = \sqrt{60^2 - 30^2} = \sqrt{2700} = 30\sqrt{3} \) km h⁻¹

Question 12. If \( 0.5\hat{i} + 0.8\hat{j} + C\hat{k} \) is a unit vector, find the value of C.
Answer:
(0.5)² + (0.8)² + C² = 1 ⇒ 0.25 + 0.64 + C² = 1 ⇒ C² = 0.11
\( C = \pm\sqrt{0.11} \) (the positive value is \( \sqrt{0.11} \approx 0.33 \))

Question 13. The rectangular components of a vector are (2, 2). The corresponding rectangular components of another vector are \( (1, \sqrt{3}) \). Find the angle between the two vectors.
Answer:
The first vector makes tan⁻¹(2/2) = 45° with the x-axis; the second makes \( \tan^{-1}\sqrt{3} \) = 60°.
Angle between them = 60° − 45° = 15°

Question 14. If \( \vec{A} = \hat{i} + \hat{j} + \hat{k} \) and \( \vec{B} = 2\hat{i} + \hat{j} \), find (a) \( \vec{A} \cdot \vec{B} \) (b) \( \vec{A} \times \vec{B} \).
Answer:
(a) \( \vec{A} \cdot \vec{B} \) = 1 × 2 + 1 × 1 + 1 × 0 = 3
(b) \( \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & 1 & 0 \end{vmatrix} = \hat{i}(0 - 1) - \hat{j}(0 - 2) + \hat{k}(1 - 2) = -\hat{i} + 2\hat{j} - \hat{k} \)

Question 15. If \( |\vec{A}| = 4 \), \( |\vec{B}| = 3 \) and θ = 60° in the figure, find (a) \( \vec{A} \cdot \vec{B} \) (b) \( |\vec{A} \times \vec{B}| \).

Answer:
(a) \( \vec{A} \cdot \vec{B} \) = AB cos 60° = 4 × 3 × ½ = 6
(b) \( |\vec{A} \times \vec{B}| \) = AB sin 60° = 12 × \( \frac{\sqrt{3}}{2} = 6\sqrt{3} \)

Question 16. Vector \( \vec{A} \) points N–E and its magnitude is 3 kg m s⁻¹. It is multiplied by the scalar λ such that λ = −4 second. Find the direction and magnitude of the new vector quantity. Does it represent the same physical quantity or not?
Answer:
\( \vec{B} = \lambda\vec{A} \) = −4 s × (3 kg m s⁻¹ towards N–E) = 12 kg m towards S–W
Its magnitude is 12 kg m, its direction is S–W (the negative scalar reverses the direction) and its unit is kg m, so it does not represent the same physical quantity.

Question 17. The angle θ between the directions of forces \( \vec{A} \) and \( \vec{B} \) is 90°, where A = 8 dyne and B = 6 dyne. If the resultant \( \vec{R} \) makes an angle α with \( \vec{A} \), then find the value of α.
Answer:
\( \tan\alpha = \frac{B\sin 90^\circ}{A + B\cos 90^\circ} = \frac{6}{8} = \frac{3}{4} \) ⇒ α = 37°

Question 18. Find the resultant of the three vectors \( \overrightarrow{OA}, \overrightarrow{OB} \) and \( \overrightarrow{OC} \), each of magnitude r, as shown in the figure.

Answer:
\( \overrightarrow{OA} + \overrightarrow{OC} \) has magnitude \( \sqrt{r^2 + r^2} = r\sqrt{2} \) and lies along \( \overrightarrow{OB} \) (at 45°).
Resultant = \( r\sqrt{2} + r = r(1 + \sqrt{2}) \), along \( \overrightarrow{OB} \).

Question 19. If \( \vec{A} = 3\hat{i} + 4\hat{j} \) and \( \vec{B} = \hat{i} + \hat{j} + 2\hat{k} \), then find the unit vector along \( \vec{A} + \vec{B} \).
Answer:
\( \vec{A} + \vec{B} = 4\hat{i} + 5\hat{j} + 2\hat{k} \), \( |\vec{A} + \vec{B}| = \sqrt{16 + 25 + 4} = \sqrt{45} \)
Unit vector \( = \frac{4\hat{i} + 5\hat{j} + 2\hat{k}}{\sqrt{45}} \)

Question 20. If \( \vec{a} = x_1\hat{i} + y_1\hat{j} \) and \( \vec{b} = x_2\hat{i} + y_2\hat{j} \), the condition that would make \( \vec{a} \) and \( \vec{b} \) parallel to each other is ______.
Answer:
For parallel vectors \( \vec{a} = \lambda\vec{b} \), i.e. x₁ = λx₂ and y₁ = λy₂.
Condition: \( \frac{x_1}{x_2} = \frac{y_1}{y_2} \)

Question 22. Two forces \( \vec{F}_1 = 2\hat{i} + 2\hat{j} \) N and \( \vec{F}_2 = 3\hat{i} + 4\hat{k} \) N are acting on a particle.
(a) Find the resultant force acting on the particle.
(b) Find the angle between \( \vec{F}_1 \) and \( \vec{F}_2 \).
(c) Find the component of force \( \vec{F}_1 \) along force \( \vec{F}_2 \).
Answer:
(a) \( \vec{F}_R = \vec{F}_1 + \vec{F}_2 = 5\hat{i} + 2\hat{j} + 4\hat{k} \) N
(b) \( \cos\theta = \frac{\vec{F}_1 \cdot \vec{F}_2}{|\vec{F}_1||\vec{F}_2|} = \frac{6}{2\sqrt{2} \times 5} = \frac{3}{5\sqrt{2}} \) ⇒ \( \theta = \cos^{-1}\left(\frac{3}{5\sqrt{2}}\right) \)
(c) \( F_1\cos\theta = \frac{\vec{F}_1 \cdot \vec{F}_2}{|\vec{F}_2|} = \frac{6}{5} \) N

Question 23. State true or false:
(i) If \( \vec{A} \) and \( \vec{B} \) are two force vectors, \( \vec{A} \cdot \vec{B} = \vec{B} \cdot \vec{A} \).
(ii) If \( \vec{A} \) and \( \vec{B} \) are two force vectors, then \( \vec{A} \times \vec{B} = \vec{B} \times \vec{A} \).
(iii) If the vector product of two non-zero vectors vanishes, the vectors are collinear.
Answer:
(i) True (the scalar product is commutative)
(ii) False (\( \vec{A} \times \vec{B} = -\vec{B} \times \vec{A} \))
(iii) True (AB sin θ = 0 ⇒ θ = 0° or 180°)

Question 24. Fill in the blanks:
(i) The scalar product of vectors \( \vec{A} = 2\hat{i} + 5\hat{k} \) and \( \vec{B} = 3\hat{j} + 5\hat{k} \) is ______.
(ii) If \( \vec{A} = 3\hat{i} + 4\hat{j} \) and \( \vec{B} = 7\hat{i} + 24\hat{j} \), then the vector having the same magnitude as \( \vec{B} \) and parallel to \( \vec{A} \) is ______.
(iii) If \( \vec{A} \parallel \vec{B} \), then \( \vec{A} \times \vec{B} \) = ______.
(iv) The magnitude of the area of the parallelogram formed by the adjacent sides of vectors \( \vec{A} = 3\hat{i} + 2\hat{j} \) and \( \vec{B} = 2\hat{i} - 2\hat{k} \) is ______.
(v) A force is represented by \( 2\hat{i} + 3\hat{j} + 6\hat{k} \). The magnitude of the force is ______.
(vi) The unit vector along vector \( \hat{i} + \hat{j} + \hat{k} \) is ______.
(vii) If \( \vec{A} \) is ______ to \( \vec{B} \), then \( \vec{A} \cdot \vec{B} = 0 \).
(viii) The vector \( \vec{A} = \hat{i} + \hat{j} \), where \( \hat{i} \) and \( \hat{j} \) are unit vectors along the x-axis and y-axis respectively, makes an angle of ______ degree with the x-axis.
(ix) If \( \vec{A} + \vec{B} + \vec{C} = \vec{0} \), then \( \vec{A} \cdot (\vec{B} \times \vec{C}) \) = ______.
Answer:
(i) 0 + 0 + 25 = 25 units
(ii) |B| = 25 and \( \hat{A} = \frac{3\hat{i} + 4\hat{j}}{5} \), so the vector is \( 25\hat{A} = 15\hat{i} + 20\hat{j} \)
(iii) Null vector
(iv) \( \vec{A} \times \vec{B} = -4\hat{i} + 6\hat{j} - 4\hat{k} \), area \( = \sqrt{16 + 36 + 16} = \sqrt{68} = 2\sqrt{17} \) units
(v) \( \sqrt{4 + 9 + 36} \) = 7 units
(vi) \( \frac{1}{\sqrt{3}}\hat{i} + \frac{1}{\sqrt{3}}\hat{j} + \frac{1}{\sqrt{3}}\hat{k} \)
(vii) Perpendicular
(viii) 45
(ix) Zero (\( \vec{A} = -(\vec{B} + \vec{C}) \) is coplanar with \( \vec{B} \) and \( \vec{C} \), so it is perpendicular to \( \vec{B} \times \vec{C} \))

Question 25. Two vectors have magnitudes 3 units and 4 units respectively. What should be the angle between them if the magnitude of the resultant is (a) 1 unit, (b) 5 units and (c) 7 units?
Answer:
R² = 9 + 16 + 24 cos θ = 25 + 24 cos θ
(a) R = 1: cos θ = −1 ⇒ θ = 180°
(b) R = 5: cos θ = 0 ⇒ θ = 90°
(c) R = 7: cos θ = 1 ⇒ θ = 0°

Question 26. When two forces of magnitude P and Q are perpendicular to each other, their resultant is of magnitude R. When they are at an angle of 180° to each other their resultant is of magnitude \( \frac{R}{\sqrt{2}} \). Find the ratio of P and Q.
Answer:
P² + Q² = R² and \( (P - Q)^2 = \frac{R^2}{2} \)
Subtracting: 2PQ = R²/2 ⇒ \( PQ = \frac{R^2}{4} \), so P² + Q² = 4PQ
With t = P/Q: t² − 4t + 1 = 0 ⇒ \( t = 2 \pm \sqrt{3} \)
P : Q = \( (2 + \sqrt{3}) : 1 \) (or its reciprocal \( (2 - \sqrt{3}) : 1 \))

Question 27. A body acted upon by 3 given forces is under equilibrium, as shown in the figure.
(a) If \( |\vec{F}_1| \) = 10 N and \( |\vec{F}_2| \) = 6 N, find the values of \( |\vec{F}_3| \) and the angle θ.
(b) Express \( \vec{F}_2 \) in unit vector form.

Answer:
(a) Horizontal: 10 sin 37° = 6 sin θ ⇒ 6 = 6 sin θ ⇒ θ = 90°
Vertical: F₃ = 10 cos 37° + 6 cos 90° = 8 N
(b) \( \vec{F}_2 \) is along the negative x-axis: \( \vec{F}_2 = -6\hat{i} \) N

Question 28. If the four forces as shown are in equilibrium, express \( \vec{F}_1 \) and \( \vec{F}_2 \) in unit vector form.

Answer:
x: −15 cos 37° + F₂ sin 30° + 10 cos 30° = 0 ⇒ −12 + F₂/2 + 5√3 = 0 ⇒ F₂ = 24 − 10√3
y: 15 sin 37° + F₂ cos 30° + 10 sin 30° − F₁ = 0 ⇒ F₁ = 9 + 5 + (24 − 10√3)(√3/2) = 12√3 − 1
\( \vec{F}_1 = -(12\sqrt{3} - 1)\hat{j} \) N and \( \vec{F}_2 = (12 - 5\sqrt{3})\hat{i} + (12\sqrt{3} - 15)\hat{j} \) N

Question 29. A particle is acted upon by the forces \( \vec{F}_1 = 2\hat{i} + a\hat{j} - 3\hat{k} \), \( \vec{F}_2 = 5\hat{i} + c\hat{j} - b\hat{k} \), \( \vec{F}_3 = b\hat{i} + 5\hat{j} - 7\hat{k} \), \( \vec{F}_4 = c\hat{i} + 6\hat{j} - a\hat{k} \). Find the values of the constants a, b, c in order that the particle will be in equilibrium.
Answer:
For equilibrium the sum of the forces is zero:
\( \hat{i} \): 2 + 5 + b + c = 0 ⇒ b + c = −7; \( \hat{j} \): a + c + 5 + 6 = 0 ⇒ a + c = −11; \( \hat{k} \): −3 − b − 7 − a = 0 ⇒ a + b = −10
Adding: 2(a + b + c) = −28 ⇒ a + b + c = −14 ⇒ a = −7, b = −3, c = −4

Question 30. A plane body has perpendicular axes OX and OY marked on it and is acted on by the following forces: 5P in the direction OY; 4P in the direction OX; 10P in the direction OA, where A is the point (3a, 4a); 15P in the direction AB, where B is the point (−a, a). Express each force in unit vector form and calculate the magnitude and direction of the sum of these forces.
Answer:
5P along OY = \( 5P\hat{j} \); 4P along OX = \( 4P\hat{i} \)
10P along OA: \( 10P \cdot \frac{3\hat{i} + 4\hat{j}}{5} = 6P\hat{i} + 8P\hat{j} \)
15P along AB: \( \overrightarrow{AB} = -4a\hat{i} - 3a\hat{j} \), so the force \( = 15P \cdot \frac{-4\hat{i} - 3\hat{j}}{5} = -12P\hat{i} - 9P\hat{j} \)
Sum \( = (4 + 6 - 12)P\hat{i} + (5 + 8 - 9)P\hat{j} = -2P\hat{i} + 4P\hat{j} \); magnitude \( = \sqrt{4 + 16}\,P = \sqrt{20}\,P \)
Direction: tan θ = 4P/(−2P) = −2, i.e. the sum makes tan⁻¹(2) ≈ 63.4° with the negative x-axis (about 116.6° with the +x axis).

Question 31. A vector \( \vec{A} \) of length 10 units makes an angle of 60° with the vector \( \vec{B} \) of length 6 units. Find the magnitude of the vector difference \( \vec{A} - \vec{B} \) and the angle it makes with vector \( \vec{A} \).
Answer:
\( |\vec{A} - \vec{B}| = \sqrt{100 + 36 - 2 \times 10 \times 6 \times \frac{1}{2}} = \sqrt{76} = 2\sqrt{19} \) units
\( (\vec{A} - \vec{B}) \cdot \vec{A} = A^2 - AB\cos 60^\circ = 100 - 30 = 70 \)
\( \cos\alpha = \frac{70}{10 \times 2\sqrt{19}} = \frac{7}{2\sqrt{19}} \) ⇒ \( \alpha = \cos^{-1}\frac{7}{2\sqrt{19}} \)

Question 32. (a) Calculate \( \vec{r} = \vec{a} - \vec{b} + \vec{c} \) where \( \vec{a} = 5\hat{i} + 4\hat{j} - 6\hat{k} \), \( \vec{b} = -2\hat{i} + 2\hat{j} + 3\hat{k} \) and \( \vec{c} = 4\hat{i} + 3\hat{j} + 2\hat{k} \).
(b) Calculate the angle between \( \vec{r} \) and the z-axis.
(c) Find the angle between \( \vec{a} \) and \( \vec{b} \).
Answer:
(a) \( \vec{r} = (5 + 2 + 4)\hat{i} + (4 - 2 + 3)\hat{j} + (-6 - 3 + 2)\hat{k} = 11\hat{i} + 5\hat{j} - 7\hat{k} \)
(b) \( |\vec{r}| = \sqrt{121 + 25 + 49} = \sqrt{195} \), \( \cos\gamma = \frac{\vec{r} \cdot \hat{k}}{|\vec{r}|} = \frac{-7}{\sqrt{195}} \) ⇒ \( \gamma = \cos^{-1}\left(\frac{-7}{\sqrt{195}}\right) \)
(c) \( \vec{a} \cdot \vec{b} = -10 + 8 - 18 = -20 \), \( |\vec{a}| = \sqrt{77} \), \( |\vec{b}| = \sqrt{17} \) ⇒ \( \theta = \cos^{-1}\left(\frac{-20}{\sqrt{1309}}\right) \)