Question

If two vectors \(\vec{A}\) and \(\vec{B}\) having equal magnitude \(R\) are inclined at an angle \(\theta\), then

• A. \(|\vec{A} - \vec{B}| = \sqrt{2} R \sin \left(\frac{\theta}{2}\right)\)
• B. \(|\vec{A} + \vec{B}| = 2 R \sin \left(\frac{\theta}{2}\right)\)
• C. \(|\vec{A} + \vec{B}| = 2 R \cos \left(\frac{\theta}{2}\right)\)
• D. \(|\vec{A} - \vec{B}| = 2 R \cos \left(\frac{\theta}{2}\right)\)

Answer 1

The Correct Option is C

To solve this problem, we need to determine the resultant magnitude of the sum or difference of two vectors \(\vec{A}\) and \(\vec{B}\) that have equal magnitudes and are inclined at an angle \(\theta\).

Let's start by considering the vector addition formula for two vectors \(\vec{A}\) and \(\vec{B}\) with equal magnitudes \(R\):

\(|\vec{A} + \vec{B}| = \sqrt{A^2 + B^2 + 2AB \cos \theta}\)

Since \(|\vec{A}| = |\vec{B}| = R\), we substitute these values into the formula:

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

Simplifying this, we get:

\(|\vec{A} + \vec{B}| = \sqrt{2R^2(1 + \cos \theta)}\)

Using the trigonometric identity \(1 + \cos \theta = 2 \cos^2 \left(\frac{\theta}{2}\right)\), we substitute to find:

\(|\vec{A} + \vec{B}| = \sqrt{2R^2 \cdot 2 \cos^2 \left(\frac{\theta}{2}\right)}\)

This reduces to:

\(|\vec{A} + \vec{B}| = 2R \cos \left(\frac{\theta}{2}\right)\)

Thus, the correct answer is:

\(|\vec{A} + \vec{B}| = 2 R \cos \left(\frac{\theta}{2}\right)\)

This matches with the given correct option: \(|\vec{A} + \vec{B}| = 2 R \cos \left(\frac{\theta}{2}\right)\).

By examining other options, we can further confirm this:

• \(|\vec{A} - \vec{B}| = \sqrt{2} R \sin \left(\frac{\theta}{2}\right)\): This doesn't fit the derived formula for \(|\vec{A} - \vec{B}|\).
• \(|\vec{A} + \vec{B}| = 2 R \sin \left(\frac{\theta}{2}\right)\): Does not match the formula derived.
• \(|\vec{A} - \vec{B}| = 2 R \cos \left(\frac{\theta}{2}\right)\): Incorrect as it reflects addition formula under the wrong identity.

Therefore, Option 3 is correct and valid.

Answer 2

The magnitude of the resultant vector \( R' \) of two vectors \( A \) and \( B \) inclined at an angle \( \theta \) is given by:

\[ R' = \sqrt{a^2 + b^2 + 2ab \cos \theta}. \]

Here \( a = b = R \), so:

\[ R' = \sqrt{R^2 + R^2 + 2R \cdot R \cos \theta} = \sqrt{2R^2 (1 + \cos \theta)}. \]

Using the identity \( 1 + \cos \theta = 2 \cos^2 \left(\frac{\theta}{2}\right) \), we get:

\[ R' = \sqrt{2R^2 \cdot 2 \cos^2 \left(\frac{\theta}{2}\right)} = 2R \cos \left(\frac{\theta}{2}\right). \]

Thus, the answer is:

\[ |A + B| = 2R \cos \left(\frac{\theta}{2}\right). \]