Problem 18 - Master Resource Book in JEE Main Physics

Question

If the magnitude of the sum of the two vectors is equal to the difference of their magnitudes, then the angle between vectors is (a) \(0^{\circ}\) (b) \(45^{\circ}\) (c) \(90^{\circ}\) (d) \(180^{\circ}\)

Step-by-Step Solution

Answer

The angle between the vectors is \( 180^{\circ} \), option (d).

Step 1: Understanding the Problem Statement

You are given two vectors, say \( \mathbf{A} \) and \( \mathbf{B} \). The magnitude of their sum \( \| \mathbf{A} + \mathbf{B} \| \) is equal to the difference of their magnitudes, which means \( \| \mathbf{A} + \mathbf{B} \| = \| \mathbf{A} \| - \| \mathbf{B} \| \). We need to find the angle between these vectors.

Step 2: Express in Mathematical Form

Use the formula for the magnitude of the sum of two vectors: \( \| \mathbf{A} + \mathbf{B} \| = \sqrt{\| \mathbf{A} \|^2 + \| \mathbf{B} \|^2 + 2 \cdot \| \mathbf{A} \| \cdot \| \mathbf{B} \| \cdot \cos \theta} \). Equate this to \( \| \mathbf{A} \| - \| \mathbf{B} \| \).

Step 3: Square Both Sides to Simplify

Square both sides of the equation: \[ (\| \mathbf{A} \| - \| \mathbf{B} \|)^2 = \| \mathbf{A} \|^2 + \| \mathbf{B} \|^2 + 2 \cdot \| \mathbf{A} \| \cdot \| \mathbf{B} \| \cdot \cos \theta \].

Step 4: Expand and Rearrange the Equation

The left side becomes \( \| \mathbf{A} \|^2 - 2 \| \mathbf{A} \| \| \mathbf{B} \| + \| \mathbf{B} \|^2 \). Equate it to the right side. This simplifies to \[ -2 \| \mathbf{A} \| \| \mathbf{B} \| = 2 \| \mathbf{A} \| \| \mathbf{B} \| \cos \theta \].

Step 5: Solve for \( \cos \theta \)

Divide both sides by \( 2 \| \mathbf{A} \| \| \mathbf{B} \| \) to isolate \( \cos \theta \): \[ -1 = \cos \theta \]. This implies that \( \theta = 180^{\circ} \).

Step 6: Conclusion from Calculations

The angle \( \theta \) calculated is \( 180^{\circ} \), which corresponds to option (d). This means the two vectors are pointing in exactly opposite directions.

Key Concepts

Magnitude of Vectors

Vectors are essential components in physics and mathematics, and understanding their magnitude is key to solving vector problems. The magnitude of a vector, denoted as \( \| \mathbf{A} \| \) for vector \( \mathbf{A} \), represents its length. It is a measurement of how much of a particular quantity is being expressed by the vector, such as force, velocity, or displacement. The magnitude provides a scalar value, stripping away the direction to focus solely on size. For a vector in two dimensions \( \mathbf{A} = \langle x, y \rangle \), the magnitude is calculated using the Pythagorean theorem:

• \( \| \mathbf{A} \| = \sqrt{x^2 + y^2} \)

In three dimensions, if \( \mathbf{A} = \langle x, y, z \rangle \), the magnitude becomes:

• \( \| \mathbf{A} \| = \sqrt{x^2 + y^2 + z^2} \)

Understanding how to calculate the magnitude helps you assess the vector's scale, enabling you to work with it correctly in equations and diagrams.

Vector Sum Formula

When dealing with multiple vectors, calculating their sum requires applying the vector sum formula. This process involves adding the components of vectors together. For instance, if we have vectors \( \mathbf{A} = \langle a_1, a_2 \rangle \) and \( \mathbf{B} = \langle b_1, b_2 \rangle \), the sum \( \mathbf{A} + \mathbf{B} \) is given by:

• \( \mathbf{A} + \mathbf{B} = \langle a_1 + b_1, a_2 + b_2 \rangle \)

This operation is essential in numerous applications, particularly when determining the resultant of two or more vectors. To compute the magnitude of the sum of vectors, the formula expands to:

• \( \| \mathbf{A} + \mathbf{B} \| = \sqrt{\| \mathbf{A} \|^2 + \| \mathbf{B} \|^2 + 2 \cdot \| \mathbf{A} \| \cdot \| \mathbf{B} \| \cdot \cos \theta} \)

This equation factors in both the individual magnitudes and the cosine of the angle between them, helping to determine how the vectors combine to form a new vector.

Cosine of Angle Between Vectors

The cosine of the angle between two vectors is a vital mathematical tool for determining the orientation of those vectors relative to each other. Given vectors \( \mathbf{A} \) and \( \mathbf{B} \), the cosine of the angle \( \theta \) between them can be derived from the dot product formula:

• \( \cos \theta = \frac{\mathbf{A} \cdot \mathbf{B}}{\| \mathbf{A} \| \cdot \| \mathbf{B} \|} \)

This relationship uses the dot product, \( \mathbf{A} \cdot \mathbf{B} \), which is calculated as the sum of the products of corresponding components of the vectors.The equation \(-1 = \cos \theta\) gives a very specific solution for \( \theta \). In the original exercise, solving \(-1 = \cos \theta\) leads to \( \theta = 180^{\circ} \), meaning the vectors are directly opposite each other, signifying a completely reverse direction. Understanding this concept is crucial in fields like physics, where direction and orientation affect the resultant of vector combinations.