Problem 29 - Master Resource Book in JEE Main Physics

Question

Which of the following statements is/are correct? (a) The magnitude of the vector \(3 \hat{i}+4 \hat{j}\) is 5 (b) A force \((3 \hat{i}+4 \hat{j}) \mathrm{N}\) acting on a particle causes a displacement \(6 \hat{j}\). The work done by the force is \(30 \mathrm{~N}\) (c) If \(\mathrm{A}\) and \(\mathrm{B}\) represent two adjacent sides of \(\mathrm{a}\) parallelogram, then \(|\mathrm{A} \times \mathrm{B}|\) give the area of that parallelogram (d) A force has magnitude \(20 \mathrm{~N}\). Its component in a direction making an angle \(60^{\circ}\) with the force is \(10 \sqrt{3} \mathrm{~N}\)

Step-by-Step Solution

Answer

Statements (a) and (c) are correct.

Step 1: Evaluate Statement (a)

The magnitude of a vector \( \mathbf{v} = a \hat{i} + b \hat{j} \) is given by \( \|\mathbf{v}\| = \sqrt{a^2 + b^2} \). For the vector \( 3 \hat{i} + 4 \hat{j} \), calculate the magnitude: \[ \|\mathbf{v}\| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \] Therefore, statement (a) is correct.

Step 2: Evaluate Statement (b)

The work done by a force \( \mathbf{F} \) causing a displacement \( \mathbf{d} \) is given by the dot product \( \mathbf{F} \cdot \mathbf{d} = |\mathbf{F}| |\mathbf{d}| \cos\theta \), where \(\theta\) is the angle between them. For \( \mathbf{F} = 3 \hat{i} + 4 \hat{j} \) and \( \mathbf{d} = 6 \hat{j} \), \[ \mathbf{F} \cdot \mathbf{d} = (3 \hat{i} + 4 \hat{j}) \cdot (0 \hat{i} + 6 \hat{j}) = 18 \] Since the angle between them is \( 90^\circ \), the work done is only caused by the \( \hat{j} \) component. Therefore, statement (b) is not correct.

Step 3: Evaluate Statement (c)

The area of a parallelogram with adjacent sides represented by vectors \( \mathbf{A} \) and \( \mathbf{B} \) is given by the magnitude of their cross product: \( |\mathbf{A} \times \mathbf{B}| \). Thus, statement (c) is correct.

Step 4: Evaluate Statement (d)

The component of a force \( \mathbf{F} \) in a direction making an angle \( \theta \) with it is given by \( |\mathbf{F}| \cos\theta \). For a force of magnitude \( 20 \mathrm{~N} \) and angle \( 60^\circ \), the component is \[ 20 \cos 60^\circ = 20 \times \frac{1}{2} = 10 \mathrm{~N} \] Since the calculated component is \( 10 \mathrm{~N} \) and not \( 10\sqrt{3} \mathrm{~N} \), statement (d) is not correct.

Key Concepts

Vector Magnitude

When dealing with vectors, one of the key aspects is finding their magnitude. The magnitude of a vector is essentially its length or size and is denoted by \( \| \mathbf{v} \| \). Suppose we have a two-dimensional vector \( \mathbf{v} = a \hat{i} + b \hat{j} \). To find its magnitude, you would use the Pythagorean formula:

• \( \| \mathbf{v} \| = \sqrt{a^2 + b^2} \)

For the vector \( 3 \hat{i} + 4 \hat{j} \), plugging in the values gives \[ \| \mathbf{v} \| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]This confirms the vector's magnitude is 5. It is akin to finding the hypotenuse in a right-angled triangle where the sides are 3 and 4 units long.

Dot Product and Work Done

The dot product is a way of multiplying two vectors that results in a scalar, rather than a vector. It's often used to calculate the work done when a force moves an object in the direction of the force. The work done is mathematically represented by the dot product:

• \( \mathbf{F} \cdot \mathbf{d} = \| \mathbf{F} \| \| \mathbf{d} \| \cos\theta \)

For the vectors \( \mathbf{F} = 3 \hat{i} + 4 \hat{j} \) and \( \mathbf{d} = 6 \hat{j} \), their dot product is simply \[ \mathbf{F} \cdot \mathbf{d} = (3 \hat{i} + 4 \hat{j}) \cdot (0 \hat{i} + 6 \hat{j}) = 18 \]In this case, the work done is entirely due to the \( \hat{j} \) components because the force and displacement share the same direction when \( \cos(0) = 1 \). The work value represents energy transferred by moving an object along a displacement by a force.

Cross Product and Area of Parallelogram

The cross product of two vectors results in another vector that is perpendicular to the plane containing the original vectors. The magnitude of this cross product reflects the area of a parallelogram formed by these vectors. If \( \mathbf{A} \) and \( \mathbf{B} \) are two adjacent sides of a parallelogram, then the area \( A \) is given by:

• \( A = \| \mathbf{A} \times \mathbf{B} \| \)

This formula highlights a unique property: the cross product not only provides a normal vector but also encloses the plane's area spanned by the two original vectors. Thus, the cross product is crucial in both physics and engineering contexts when dealing with vector spaces.