H C Verma Solution Part I Chapter 2 | Physics and Mathematics

Table of Contents

QUESTIONS FOR SHORT ANSWER

1. Is a vector necessarily changed if it is rotated through an angle ?
Yes.

Explanation:
A vector depends on magnitude and direction. Rotation changes direction, so the vector changes. Exceptions: zero vector or rotation by multiples of $360^\circ$.

2. Is it possible to add two vectors of unequal magnitudes and get zero? Is it possible to add three vectors of equal magnitudes and get zero?
No for the first part; Yes for the second part.

Explanation:
Two vectors give zero only if equal and opposite. Unequal magnitudes cannot cancel. Three equal vectors can give zero if arranged at $120^\circ$ forming a closed triangle.

Explanation: Think of it like a tug of war. For two forces or vectors to give a zero resultant, two conditions must be satisfied.

Direction: The vectors must act in exactly opposite directions. If they act at any angle, there will be a sideways motion.

Magnitude: The vectors must have equal magnitude. If one is even slightly larger, there will be a net force in the direction of the stronger vector.

Mathematically, for vectors $$\vec{A}$$ and $$\vec{B}$$, the resultant is zero only when:

$$ \vec{A} + \vec{B} = \vec{0} $$

This implies:

$$ \vec{A} = -\vec{B} $$

Thus, both vectors must have same magnitude and opposite direction.

3. Does the phrase “direction of zero vector” have physical significance? Discuss it in terms of velocity, force etc.

Answer: No, the phrase “direction of zero vector” does not have any physical significance.

Explanation: A zero vector (or null vector) is defined as a vector having zero magnitude and an indeterminate or arbitrary direction.

In mathematical form, a zero vector is written as:

$$ \vec{0} $$

In physical terms, if a body is at rest, its velocity is represented by a zero vector:

$$ \vec{v} = \vec{0} $$

It is meaningless to specify a direction for its motion because no motion is occurring.

Similarly, if the net force acting on a particle is zero:

$$ \sum \vec{F} = \vec{0} $$

the particle is in equilibrium and does not accelerate in any specific direction.

Since a vector’s direction is only defined when there is a non-zero displacement, flow, or force, the direction of a zero vector remains purely a mathematical convention without a unique physical orientation.

4. Can you add three unit vectors to get a unit vector? Does your answer change if two unit vectors are along the coordinate axes ?

Answer: Yes, it is possible to add three unit vectors to obtain a resultant that is also a unit vector.

Explanation: A unit vector has a magnitude of exactly one.

$$ |\vec{a}| = 1 $$

For the sum of three such vectors to result in another unit vector, their resultant must also satisfy:

$$ |\vec{R}| = 1 $$

One simple way to achieve this is by choosing two unit vectors in opposite directions so that they cancel each other.

Let the three unit vectors be:

$$ \hat{i}, \quad -\hat{i}, \quad \hat{j} $$

Their resultant is:

$$ \vec{R} = \hat{i} – \hat{i} + \hat{j} $$

$$ \vec{R} = \hat{j} $$

Now, the magnitude of the resultant is:

$$ |\vec{R}| = |\hat{j}| = 1 $$

Thus, three unit vectors can be added to obtain another unit vector.

5. Can we have physical quantities having magnitude and direction which are not vectors ?

Ans: Yes, such physical quantities exist.

Explanation: A vector quantity must satisfy the law of vector addition (triangle law or parallelogram law). Some physical quantities have magnitude and direction but do not obey these laws, so they are not vectors. Examples include:

Electric current: It has magnitude and a specified direction of flow, but currents do not add according to vector rules.
Magnetic flux: It has magnitude and direction through a surface, but it does not follow vector addition.
Angle of rotation: It has magnitude and direction (clockwise or anticlockwise), but it is not treated as a vector.

Thus, while these quantities possess both magnitude and direction, they are classified as directed quantities rather than true vectors.

6. Which of the following two statements is more appropriate ?

(a) Two forces are added using triangle rule because force is a vector quantity. (b) Force is a vector quantity because two forces are added using triangle rule.

Answer: (a) Two forces are added using triangle rule because force is a vector quantity.

Explanation: A force is a vector quantity since it has both magnitude and direction and obeys the law of vector addition. The triangle rule is a method of adding two vectors, and it applies to forces because they are vectors. Statement (b) is incorrect because the nature of force (being a vector) is not determined by the rule of addition; rather, the rule of addition applies because force is already a vector quantity.

7. Can you add two vectors representing physical quantities having different dimensions ? Can you multiply two vectors representing physical quantities having different dimensions ?

Answer: You cannot add two vectors representing physical quantities having different dimensions, but you can multiply them.

Explanation:

Addition of vectors is only possible when the vectors represent the same physical quantity and have the same dimensions. For example, you can add two forces or two velocities, but you cannot add a force and a velocity, because they have different dimensions.
Multiplication of vectors is possible even if they represent different physical quantities. For instance, multiplying force and velocity gives power, multiplying velocity and momentum gives another physical quantity. Multiplication can be done as dot product (scalar result) or cross product (vector result).

Thus, addition requires same dimensions, while multiplication can involve different dimensions and often leads to meaningful new physical quantities.

8. Can a vector have zero component along a line and still have nonzero magnitude ?

Answer: Yes, a vector can have zero component along a line and still have nonzero magnitude.

Explanation: A component of a vector along a line represents the projection of the vector in that direction. If the vector is perpendicular to the line, its component along that line is zero. However, the vector itself may still have a nonzero magnitude because it exists in a different direction. For example, a vector along the y-axis has zero component along the x-axis, but its magnitude is not zero. Thus, a vector can have zero component along a line while still possessing a nonzero magnitude.

9. Let $ \epsilon_1 $ and $ \epsilon_2 $ be the angles made by $ \vec{A} $ and $ -\vec{A} $ with the positive X-axis. Show that $ \tan \epsilon_1 = \tan \epsilon_2 $. Thus, giving $ \tan \epsilon $ does not uniquely determine the direction of $ \vec{A} $.

Answer: Result verified.

Explanation: The vector $ -\vec{A} $ is directed exactly opposite to $ \vec{A} $, meaning it is rotated by $ 180^\circ $. If $ \vec{A} $ makes an angle $ \epsilon_1 $ with the X-axis, then $ -\vec{A} $ makes an angle $ \epsilon_2 = \epsilon_1 + 180^\circ $. Using the identity $ \tan(\theta + 180^\circ) = \tan \theta $, we get $ \tan \epsilon_1 = \tan \epsilon_2 $. This shows that the tangent of an angle gives the same value for two opposite directions, so it does not uniquely determine the direction of a vector.

10. Is the vector sum of the unit vectors $ \hat{i} $ and $ \hat{j} $ a unit vector ? If no, can you multiply this sum by a scalar number to get a unit vector ?

Answer: No, the vector sum of the unit vectors $ \hat{i} $ and $ \hat{j} $ is not a unit vector. Yes, it can be multiplied by a scalar to get a unit vector.

Explanation: The unit vectors are $ \hat{i} = (1, 0) $ and $ \hat{j} = (0, 1) $. Their sum is $ \hat{i} + \hat{j} = (1, 1) $. The magnitude of this vector is $ \sqrt{1^2 + 1^2} = \sqrt{2} $, which is not equal to 1, so it is not a unit vector. To convert it into a unit vector, divide by its magnitude: $ \frac{\hat{i} + \hat{j}}{\sqrt{2}} = \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) $. The magnitude of this new vector is $ \sqrt{\frac{1}{2} + \frac{1}{2}} = 1 $. Hence, multiplying by the scalar $ \frac{1}{\sqrt{2}} $ gives a unit vector.

11. Let $ \vec{A} = 3\hat{i} + 4\hat{j} $. Write four vector $ \vec{B} $ such that $ \vec{A} \ne \vec{B} $ but $ |\vec{A}| = |\vec{B}| $.

Answer: $ \vec{B}_1 = 5\hat{i} $, $ \vec{B}_2 = 5\hat{j} $, $ \vec{B}_3 = 4\hat{i} + 3\hat{j} $, $ \vec{B}_4 = -3\hat{i} – 4\hat{j} $.

Explanation: The given vector $ \vec{A} = 3\hat{i} + 4\hat{j} $ has magnitude $ \sqrt{3^2 + 4^2} = 5 $. To find vectors $ \vec{B} $ such that $ \vec{A} \ne \vec{B} $ but $ |\vec{A}| = |\vec{B}| $, we need vectors having the same magnitude but different components. The vectors $ 5\hat{i} $, $ 5\hat{j} $, $ 4\hat{i} + 3\hat{j} $, and $ -3\hat{i} – 4\hat{j} $ all have magnitude 5, but are different in direction from $ \vec{A} $. Hence, they satisfy the required condition.

12. Can you have $ \vec{A} \times \vec{B} = \vec{A} \cdot \vec{B} $ with $ \vec{A} \ne 0 $ and $ \vec{B} \ne 0 $ ? What if one of the two vectors is zero ?

Answer: Yes.

Explanation: The magnitude of cross product is $ AB \sin\theta $ and the dot product is $ AB \cos\theta $. For them to be equal in magnitude, we must have $ \sin\theta = \cos\theta $, which occurs when $ \theta = 45^\circ $. Thus, for nonzero vectors, the equality holds in magnitude when the angle between them is 45^\circ. If one of the vectors is zero, then both the dot product and cross product become zero, so the equality is satisfied for any angle.

13. If ( A × B = 0 ), can you say that (a) A = B, (b) A ≠ B ?

Answer: You cannot conclude that A = B. The correct statement is that A and B are parallel (or anti-parallel), which means they may or may not be equal.

Explanation: The cross product of two vectors is given by: A × B = AB sinθ, where θ is the angle between them.

If A × B = 0, then either A = 0, B = 0, or sinθ = 0.
sinθ = 0 means θ = 0° or 180°, so the vectors are parallel or anti-parallel.
This does not necessarily mean A = B. For example, A = 2i and B = 3i are parallel but not equal. Similarly, A = i and B = -i are anti-parallel.

Thus, the condition A × B = 0 only tells us that the vectors are parallel or anti-parallel, not that they are equal.

14. Let A = 5i – 4j and B = –7.5i + 6j. Do we have B = kA ? Can we say B / A = k ?

Answer: Yes, we have B = kA with k = –1.5. But we cannot say B / A = k in the usual sense of division of vectors.

Explanation: Given: A = 5i – 4j B = –7.5i + 6j

Now, check if B = kA: For the i-component: –7.5 = k × 5 ⇒ k = –7.5 / 5 = –1.5 For the j-component: 6 = k × (–4) ⇒ k = 6 / (–4) = –1.5

Since both give the same value, k = –1.5, we conclude that B = –1.5A.

However, the expression B / A = k is not valid, because division of vectors is not defined in vector algebra. We can only say that B is a scalar multiple of A, not that one vector is divided by another.

Thus, B = kA is true with k = –1.5, but B / A = k is not a meaningful statement in vector mathematics.

OBJECTIVE I

1. A vector is not changed if:

(a) it is rotated through an arbitrary angle

(b) it is multiplied by an arbitrary scalar

(d) it is slid parallel to itself

Answer: (d)

Explanation: A vector is characterized by its magnitude and direction. Sliding a vector parallel to itself, also known as parallel translation, ensures that both its length and orientation remain constant. Since these core properties do not change, the vector remains the same.

2. Which of the sets given below may represent the magnitudes of three vectors adding to zero ?

(a) 2, 4, 8 (b) 4, 8, 16 (c) 1, 2, 1 (d) 0.5, 1, 2

Answer: (c) 1, 2, 1

Explanation: For three vectors to add to zero, their magnitudes must satisfy the triangle inequality: the sum of any two sides must equal the third side.

In (a) and (b), the largest number is greater than the sum of the other two, so they cannot form a closed triangle.
In (d), 0.5 + 1 = 1.5, which is less than 2, so they cannot form a closed triangle.
In (c), 1 + 1 = 2, which equals the third side. This satisfies the condition, meaning the three vectors can form a closed triangle and hence add to zero.

3. The resultant of A and B makes an angle alpha with A and beta with B,

(a) alpha < beta

(b) alpha < beta if A < B

(d) alpha < beta if A = B

Answer: (c)

Explanation: The resultant of two vectors always inclines more toward the vector with the larger magnitude. If magnitude A > B, the resultant will be closer to A, meaning the angle alpha (with A) will be smaller than the angle beta (with B).

4. The component of a vector is

(a) always less than its magnitude

(b) always greater than its magnitude

(d) none of these.

Answer: (d)

Explanation: The component of a vector along any direction is given by the magnitude multiplied by the cosine of the angle between them. This value can be less than the magnitude (if the angle is non-zero) or equal to the magnitude (if the angle is zero), but it can never be greater than the magnitude.

5. A vector A points vertically upward and B points towards north. The vector product A x B is

(a) along west

(b) along east

(d) vertically downward.

Answer: (a)

Explanation: Using the right-hand thumb rule for cross product, point your fingers vertically upward (A) and curl them towards north (B). Your thumb points towards the west, which gives the direction of the resultant vector.

6. The radius of a circle is r. The area of the circle is A = $$\pi r^2$$. The rate of change of area with respect to its radius is

(a) $$2\pi r$$

(b) $$2\pi$$

(d) $$\pi$$

Answer: (a)

Explanation: The rate of change of one quantity with respect to another is found using differentiation. Differentiating the area formula $$A = \pi r^2$$ with respect to $$r$$ gives $$\frac{dA}{dr} = 2\pi r$$, which represents the circumference of the circle.

OBJECTIVE II

1. A situation may be described by using different coordinate systems having different orientations of the axes.

(a) the scalar values will be same in all systems

(b) the vector values will be same in all systems

(d) the magnitude of a vector will be same in all systems

Answer: (a), (b), (d)

Explanation: Physical quantities like scalars and vectors exist independently of the coordinate system, so their values and magnitudes remain unchanged. However, the components of a vector depend on the orientation of axes, so they change when the coordinate system is rotated.

2. The x-component of the resultant of several vectors

(a) is 840 N

(b) is greater than the sum of x-components of the vectors

(d) may be equal to the sum of magnitudes of the vectors

Answer: (c), (d)

Explanation: The x-component of the resultant is equal to the algebraic sum of individual x-components. Since a component is always less than or equal to its magnitude, it can be smaller than the sum of magnitudes and becomes equal only when all vectors lie along the x-axis.

3. The magnitude of the sum of two vectors A and B is less than the magnitude of A – B if

(a) A and B are perpendicular

(b) A and B make an acute angle

(d) A and B are in opposite directions

Answer: (c), (d)

Explanation: The magnitude of the sum depends on the angle between vectors. If the angle is obtuse, the vectors oppose each other, making the sum smaller than the difference. In opposite directions, the sum is minimum and the difference is maximum.

4. Which of the following quantities are independent of the choice of orientation of the coordinate axes?

(a) A + B

(b) A_x + B_x

(d) Angle between A and B

Answer: (a), (c), (d)

Explanation: The vector sum, its magnitude, and the angle between vectors are independent of the orientation of axes. However, components depend on the coordinate system and therefore change when the axes are rotated.

5. The magnitude of a vector product of two vectors A and B may be

(a) greater than AB

(b) equal to AB

(d) equal to zero

Answer: (b), (c), (d)

Explanation: The magnitude of a vector product depends on the angle between vectors. It can be zero when vectors are parallel, equal to AB when vectors are perpendicular, or less than AB for other angles. It can never be greater than AB.

EXERCISES

1. A vector A makes an angle of 20° and B makes an angle of 110° with the X-axis. The magnitudes of these vectors are 3 m and 4 m respectively. Find the resultant.

Answer: The resultant vector has magnitude 5 m and makes an angle of approximately 63.4° with the X-axis.

Explanation: Let A = 3 m at 20° B = 4 m at 110°

Resolve into components:

Ax = 3 cos 20° = 2.82
Ay = 3 sin 20° = 1.03
Bx = 4 cos 110° = –1.37
By = 4 sin 110° = 3.76

Now, Rx = Ax + Bx = 2.82 – 1.37 = 1.45 Ry = Ay + By = 1.03 + 3.76 = 4.79

Magnitude of resultant: R = √(Rx² + Ry²) = √(1.45² + 4.79²) = √(2.10 + 22.94) = √25.04 ≈ 5 m

Direction of resultant: θ = tan⁻¹(Ry / Rx) = tan⁻¹(4.79 / 1.45) ≈ tan⁻¹(3.30) ≈ 73.4°

So, the resultant vector has magnitude 5 m and is inclined at 73.4° to the X-axis.

2. Let $ \vec{A} $ and $ \vec{B} $ be the two vectors of magnitude 10 units each. If they are inclined to the X-axis at angles 30° and 60° respectively, find the resultant.

Answer: 19.3 units at 45° with the X-axis.

Explanation: Resolve each vector into components. For $ \vec{A} $, $ A_x = 10\cos30^\circ = 5\sqrt{3} $ and $ A_y = 10\sin30^\circ = 5 $.

For $ \vec{B} $, $ B_x = 10\cos60^\circ = 5 $ and $ B_y = 10\sin60^\circ = 5\sqrt{3} $.

Adding components gives $ R_x = 5\sqrt{3} + 5 $ and $ R_y = 5 + 5\sqrt{3} $, so $ R_x = R_y $.

Hence, the magnitude of the resultant is $ R = \sqrt{R_x^2 + R_y^2} \approx 19.3 $ units.

The direction is given by $ \theta = \tan^{-1}(R_y/R_x) = 45^\circ $. Thus, the resultant makes an angle of 45° with the X-axis.

3. Add vectors $ \vec{A} $, $ \vec{B} $ and $ \vec{C} $ each having magnitude of 100 units and inclined to the X-axis at angles 45°, 135° and 315° respectively.

Answer: 100 units at 45° with the X-axis.

Explanation: Resolve each vector into components. For $ \vec{A} $, $ A_x = 100\cos45^\circ $ and $ A_y = 100\sin45^\circ $.

For $ \vec{B} $, $ B_x = 100\cos135^\circ $ and $ B_y = 100\sin135^\circ $.

For $ \vec{C} $, $ C_x = 100\cos315^\circ $ and $ C_y = 100\sin315^\circ $.

Adding components gives $ R_x = 70.71 $ and $ R_y = 70.71 $.

The magnitude of the resultant is $ R = \sqrt{R_x^2 + R_y^2} = 100 $.

The direction is $ \theta = \tan^{-1}(R_y/R_x) = 45^\circ $. Hence, the resultant has magnitude 100 units and is inclined at 45° to the X-axis.

4. Let $ \vec{a} = 4\hat{i} + 3\hat{j} $ and $ \vec{b} = 3\hat{i} + 4\hat{j} $. (a) Find the magnitudes of $ \vec{a} $, (b) $ \vec{b} $, (c) $ \vec{a} + \vec{b} $ and (d) $ \vec{a} – \vec{b} $.

Answer: (a) 5 (b) 5 (c) $ 7\sqrt{2} $ (d) $ \sqrt{2} $

Explanation: The magnitude of a vector is given by $ \sqrt{x^2 + y^2} $.

For $ \vec{a} = 4\hat{i} + 3\hat{j} $, $ |\vec{a}| = \sqrt{4^2 + 3^2} = 5 $.

For $ \vec{b} = 3\hat{i} + 4\hat{j} $, $ |\vec{b}| = \sqrt{3^2 + 4^2} = 5 $.

The sum $ \vec{a} + \vec{b} = 7\hat{i} + 7\hat{j} $ has magnitude $ \sqrt{7^2 + 7^2} = 7\sqrt{2} $. The difference $ \vec{a} – \vec{b} = \hat{i} – \hat{j} $ has magnitude $ \sqrt{1^2 + (-1)^2} = \sqrt{2} $.

Thus, all required magnitudes are obtained.

5. Refer to figure (2-E1). Find (a) the magnitude, (b) x and y components and (c) the angle with the X-axis of the resultant of OA, BC and DE.

Answer: (a) 1.6 m (b) 0.98 m and 1.3 m respectively (c) $ \tan^{-1}(1.32) $

Explanation: Using vector resolution, each vector is broken into its x and y components and then added. The total components are $ R_x \approx 0.98 $ m and $ R_y \approx 1.3 $ m. The magnitude of the resultant is calculated using $ \sqrt{R_x^2 + R_y^2} $, giving approximately 1.6 m. The direction with the X-axis is found using $ \tan^{-1}(R_y/R_x) $, which gives $ \tan^{-1}(1.3/0.98) \approx \tan^{-1}(1.32) $.

6. Two vectors have magnitudes 3 unit and 4 unit respectively. What should be the angle between them if the magnitude of the resultant is (a) 1 unit, (b) 5 unit and (c) 7 unit.

Answer: (a) $ 180^\circ $ (b) $ 90^\circ $ (c) $ 0^\circ $.

Explanation: Use $ R = \sqrt{A^2 + B^2 + 2AB\cos\theta} $.
(a) $ 1 = \sqrt{9+16+24\cos\theta} \Rightarrow \cos\theta=-1 \Rightarrow 180^\circ $.
(b) $ 25=9+16+24\cos\theta \Rightarrow \cos\theta=0 \Rightarrow 90^\circ $.
(c) $ 49=9+16+24\cos\theta \Rightarrow \cos\theta=1 \Rightarrow 0^\circ $.

7. A spy report about a suspected car reads as follows. “The car moved 2.00 km towards east, made a perpendicular left turn, ran for 500 m, made a perpendicular right turn, ran for 4.00 km and stopped”. Find the displacement of the car.

Answer: $ 6.02 \, \text{km}, \tan^{-1}(1/12) $.

Explanation: Net vector $ = 6\hat{i} + 0.5\hat{j} $.
Magnitude $ = \sqrt{6^2 + 0.5^2} \approx 6.02 \, \text{km} $.
Direction $ = \tan^{-1}\left(\frac{0.5}{6}\right) = \tan^{-1}(1/12) $.

8. A carrom board (4 ft × 4 ft square) has the queen at the centre. The queen, hit by the striker moves to the front edge, rebounds and goes in the hole behind the striking line. Find the magnitude of displacement of the queen (a) from the centre to the front edge, (b) from the front edge to the hole and (c) from the centre to the hole.

Answer: (a) $ \frac{2}{3}\sqrt{10} \, \text{ft} $ (b) $ \frac{4}{3}\sqrt{10} \, \text{ft} $ (c) $ 2\sqrt{2} \, \text{ft} $.

Explanation: Using geometry of reflection and coordinates:
Total displacement from $ (2,2) $ to $ (0,0) $ gives $ 2\sqrt{2} $.
Intermediate distances follow from similar triangles.

9. A mosquito net over a 7 ft × 4 ft bed is 3 ft high. The net has a hole at one corner of the bed through which a mosquito enters the net. It flies and sits at the diagonally opposite upper corner of the net. (a) Find the magnitude of the displacement of the mosquito. (b) Taking the hole as the origin, the length of the bed as the X-axis, its width as the Y-axis, and vertically up as the Z-axis, write the components of the displacement vector.

Answer: (a) $ \sqrt{74} \, \text{ft} $ (b) $ (7,4,3) $.

Explanation: Displacement from $ (0,0,0) $ to $ (7,4,3) $.
Magnitude $ = \sqrt{7^2 + 4^2 + 3^2} = \sqrt{74} $.

10. Suppose $ \vec{a} $ has magnitude $ 4.5 $ due north. Find $ 3\vec{a} $ and $ -4\vec{a} $.

Answer: (a) $ 13.5 $ due north (b) $ 18 $ due south.

Explanation: Scalar multiplication changes magnitude and possibly direction.
$ 3\vec{a} \Rightarrow 13.5 $ north,
$ -4\vec{a} \Rightarrow 18 $ south.

11. Two vectors have magnitudes 2 m and 3 m. The angle between them is 60°. Find (a) the scalar product of the two vectors, (b) the magnitude of their vector product.

Answer: (a) 3 m² (b) $ 3\sqrt{3} $ m²

Explanation: The scalar product is given by $ \vec{A} \cdot \vec{B} = AB \cos\theta $. Substituting the values gives $ (2)(3)\cos60^\circ = 3 $ m². The magnitude of vector product is given by $ |\vec{A} \times \vec{B}| = AB \sin\theta $. Substituting the values gives $ (2)(3)\sin60^\circ = 3\sqrt{3} $ m². Thus, the required results are obtained.

12. Let $ A_1 A_2 A_3 A_4 A_5 A_6 A_1 $ be a regular hexagon. Write the x-components of the vectors represented by the six sides taken in order. Use the fact that the resultant of these six vectors is zero, to prove that $ \cos 0 + \cos \frac{\pi}{3} + \cos \frac{2\pi}{3} + \cos \frac{3\pi}{3} + \cos \frac{4\pi}{3} + \cos \frac{5\pi}{3} = 0 $ and verify using known cosine values.

Answer: Result verified.

Explanation: In a regular hexagon, the six sides are equally inclined at angles $ 0, \frac{\pi}{3}, \frac{2\pi}{3}, \pi, \frac{4\pi}{3}, \frac{5\pi}{3} $ with the X-axis.

If each side has magnitude $ a $, their x-components are proportional to $ \cos $ of these angles. Since the hexagon forms a closed figure, the resultant of all six vectors is zero, so the sum of their x-components must also be zero.

Hence, $ \cos 0 + \cos \frac{\pi}{3} + \cos \frac{2\pi}{3} + \cos \pi + \cos \frac{4\pi}{3} + \cos \frac{5\pi}{3} = 0 $. Substituting known values $ 1, \frac{1}{2}, -\frac{1}{2}, -1, -\frac{1}{2}, \frac{1}{2} $, their sum becomes zero, verifying the result.

13. Let $ \vec{a} = 2\hat{i} + 3\hat{j} + 4\hat{k} $ and $ \vec{b} = 3\hat{i} + 4\hat{j} + 5\hat{k} $. Find the angle between them.

Answer: $ \cos^{-1}\left(\frac{38}{\sqrt{1450}}\right) $ (approximately 3°)

Explanation: The angle between two vectors is given by the dot product formula $ \cos\theta = \frac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|} $. The dot product is $ (2)(3) + (3)(4) + (4)(5) = 38 $. The magnitudes are $ |\vec{a}| = \sqrt{29} $ and $ |\vec{b}| = \sqrt{50} $. Substituting these gives $ \cos\theta = \frac{38}{\sqrt{29}\sqrt{50}} = \frac{38}{\sqrt{1450}} $. Evaluating this gives an angle of approximately 3°, showing the vectors are nearly parallel.

14. Prove that $ \vec{A} \cdot (\vec{A} \times \vec{B}) = 0 $.

Explanation: The cross product $ \vec{A} \times \vec{B} $ produces a vector that is perpendicular to both $ \vec{A} $ and $ \vec{B} $.

Hence, the angle between $ \vec{A} $ and $ \vec{A} \times \vec{B} $ is $ 90^\circ $.

Using the dot product formula, $ \vec{A} \cdot (\vec{A} \times \vec{B}) = |\vec{A}| \, |\vec{A} \times \vec{B}| \cos 90^\circ $.

Since $ \cos 90^\circ = 0 $, the result becomes zero. Thus, $ \vec{A} \cdot (\vec{A} \times \vec{B}) = 0 $.

15. If $ \vec{A} = 2\hat{i} + 3\hat{j} + 4\hat{k} $ and $ \vec{B} = 4\hat{i} + 3\hat{j} + 2\hat{k} $, find $ \vec{A} \times \vec{B} $.

Answer: $ -6\hat{i} + 12\hat{j} – 6\hat{k} $

Explanation: The cross product is evaluated using the determinant method. Expanding gives $ \hat{i}(3\times2 – 4\times3) – \hat{j}(2\times2 – 4\times4) + \hat{k}(2\times3 – 3\times4) $. This simplifies to $ \hat{i}(6 – 12) – \hat{j}(4 – 16) + \hat{k}(6 – 12) $, resulting in $ -6\hat{i} + 12\hat{j} – 6\hat{k} $.

16. If $ \vec{A} $, $ \vec{B} $, $ \vec{C} $ are mutually perpendicular, show that $ \vec{C} \times (\vec{A} \times \vec{B}) = 0 $. Is the converse true ?

Answer: No.

Explanation: If $ \vec{A} $ and $ \vec{B} $ are perpendicular, then $ \vec{A} \times \vec{B} $ is a vector perpendicular to both, hence it is parallel to $ \vec{C} $. The cross product of parallel vectors is zero, so $ \vec{C} \times (\vec{A} \times \vec{B}) = 0 $. The converse is not true because the result can be zero even if vectors are not mutually perpendicular, for example when any vector is zero or when $ \vec{C} $ is simply parallel to $ \vec{A} \times \vec{B} $.

17. A particle moves on a given straight line with a constant speed v. At a certain time it is at a point P on its straight line path. O is a fixed point. Show that OP × v is independent of the position P.

Solution:

Step 1: Represent the motion mathematically
Let the straight line be defined by a direction vector n̂. The velocity of the particle is constant:
v = v n̂
where v is the constant speed.

At time t, the position of the particle relative to O is:
OP = OP₀ + (vt) n̂
where OP₀ is the initial position vector at t = 0.

Step 2: Compute the cross product
OP × v = (OP₀ + vt n̂) × (v n̂)

Expanding:
OP × v = OP₀ × (v n̂) + (vt n̂) × (v n̂)

Step 3: Simplify
Since n̂ × n̂ = 0:
(vt n̂) × (v n̂) = 0

Therefore:
OP × v = OP₀ × (v n̂)

Step 4: Interpretation
This expression depends only on OP₀ and the constant velocity v. It does not depend on the current position P.

Answer: OP × v is independent of the position P.

Explanation: The term involving position P vanishes because the cross product of parallel vectors is zero. Hence, OP × v remains constant and depends only on the initial position and velocity.

18. The force on a charged particle due to electric and magnetic fields is given by F = qE + q(v × B). Suppose E is along the X-axis and B along the Y-axis. In what direction and with what minimum speed v should a positively charged particle be sent so that the net force on it is zero?

Solution:

Step 1: Condition for zero force
For net force to be zero:
F = 0 ⇒ E + (v × B) = 0
So,
v × B = −E

Step 2: Assign directions
E = E î (along X-axis)
B = B ĵ (along Y-axis)

Step 3: Assume velocity components
Let v = vₓ î + vᵧ ĵ + v_z k̂
Then,
v × B = (vₓ î + vᵧ ĵ + v_z k̂) × (B ĵ)

Expanding:
vₓ î × B ĵ = vₓ B k̂
vᵧ ĵ × B ĵ = 0
v_z k̂ × B ĵ = −v_z B î

So,
v × B = vₓ B k̂ − v_z B î

Step 4: Compare with −E î
vₓ B k̂ − v_z B î = −E î

This gives:
vₓ = 0
−v_z B = −E ⇒ v_z = E/B

Step 5: Final result
The particle must move along the positive Z-axis with speed:

v = E/B

Answer: Direction: Along positive Z-axis; Speed: v = E/B

Explanation: The magnetic force must exactly cancel the electric force. This happens when velocity is perpendicular to both E and B, and its magnitude is such that qvB = qE, giving v = E/B.

19. Give an example for which A · B = C · B but A ≠ C.

Answer: Example provided.

Explanation: Consider vector B = î. Let A = î + ĵ and C = î + k̂. Then,

A · B = (î + ĵ) · î = 1
C · B = (î + k̂) · î = 1

Thus, A · B = C · B, but A ≠ C because A and C have different components. The scalar products are equal because the components of A and C along B are identical, even though the vectors themselves are different.

20. Draw a graph from the following data. Draw tangents at x = 2, 4, 6 and 8. Find the slopes of these tangents. Verify that the curve drawn is y = 2x² and the slope of tangent is tanθ = dy/dx = 4x.

Solution:

Step 1: Verify the given data
For different values of x:
x = 1 → y = 2(1)² = 2
x = 2 → y = 2(2)² = 8
x = 3 → y = 2(3)² = 18
…
Thus, all values satisfy y = 2x².

Step 2: Equation of the curve

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Step 3: Find slope of tangent
The slope of tangent at any point is given by:

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Step 4: Calculate slopes at given points
At x = 2 → slope = 8
At x = 4 → slope = 16
At x = 6 → slope = 24
At x = 8 → slope = 32

Answer: Result verified.

Explanation: The given data follows the equation y = 2x². The slope of the tangent at any point is given by dy/dx = 4x. The calculated slopes at x = 2, 4, 6, and 8 are 8, 16, 24, and 32 respectively, which confirms the result.

21. A curve is represented by y = sin x. If x is changed from π/3 to (π/3 + π/100), find approximately the change in y.

Solution:

Step 1: Use approximation formula
For small change in x,
Δy ≈ (dy/dx) Δx

Step 2: Derivative of the function

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Step 3: Evaluate at x = π/3
cos(π/3) = 0.5

Step 4: Calculate change in x
Δx = π/100 ≈ 0.0314

Step 5: Compute change in y
Δy ≈ 0.5 × 0.0314 = 0.0157

Answer: 0.0157

Explanation: For small changes, Δy ≈ (dy/dx)Δx. Here dy/dx = cos x and at x = π/3, cos x = 0.5. Multiplying by Δx = π/100 gives Δy ≈ 0.0157.

22. The electric current in a charging R–C circuit is given by i = i₀ e^(−t/RC) where i₀, R and C are constant parameters of the circuit and t is time. Find the rate of change of current at (a) t = 0, (b) t = RC, (c) t = 10RC.

Solution:

Step 1: Differentiate the current

Step 2: Evaluate at given times

(a) At t = 0:
di/dt = −i₀ / RC

(b) At t = RC:
di/dt = −i₀ / (RC e)

Answer: (a) −i₀/RC (b) −i₀/(RCe) (c) −i₀/(RCe¹⁰)

Explanation: The rate of change of current is given by di/dt. Differentiating i = i₀ e^(−t/RC) gives di/dt = (−i₀/RC)e^(−t/RC). Substituting t = 0, RC and 10RC gives the required results.

Licchavi Lyceum

QUESTIONS FOR SHORT ANSWER

OBJECTIVE I

OBJECTIVE II

EXERCISES