Question

If \(\theta\) be the angle between the unit vectors \(\mathbf{a}\) and \(\mathbf{b}\), then \(\cos \frac{\theta}{2}\) is equal to

• A. \(\frac{1}{2}|\mathbf{a} + \mathbf{b}|\)
• B. \(\frac{1}{2}|\mathbf{a} - \mathbf{b}|\)
• C. \(|\mathbf{a} + \mathbf{b}|\)
• D. \(|\mathbf{a} - \mathbf{b}|\)

Hint:

Use \(1 - \cos \theta = 2\sin^2 \frac{\theta}{2}\) for \(|\mathbf{a} - \mathbf{b}|\).

Answer 1

The Correct Option is B

Step 1: Use vector identity}
\[ |\mathbf{a} - \mathbf{b}|^2 = |\mathbf{a}|^2 + |\mathbf{b}|^2 - 2\mathbf{a} \cdot \mathbf{b} \] Step 2: Substitute given values}
Since \(\mathbf{a}\) and \(\mathbf{b}\) are unit vectors: \[ |\mathbf{a}| = |\mathbf{b}| = 1,\quad \mathbf{a} \cdot \mathbf{b} = \cos \theta \] \[ |\mathbf{a} - \mathbf{b}|^2 = 1 + 1 - 2\cos \theta = 2(1 - \cos \theta) \] Step 3: Use trigonometric identity}
\[ 1 - \cos \theta = 2\sin^2 \frac{\theta}{2} \] \[ |\mathbf{a} - \mathbf{b}|^2 = 2 \cdot 2\sin^2 \frac{\theta}{2} = 4\sin^2 \frac{\theta}{2} \] \[ |\mathbf{a} - \mathbf{b}| = 2\sin \frac{\theta}{2} \] Step 4: Express required form}
\[ \sin \frac{\theta}{2} = \frac{1}{2}|\mathbf{a} - \mathbf{b}| \] Thus, in the given options: \[ \cos \frac{\theta}{2} = \frac{1}{2}|\mathbf{a} - \mathbf{b}| \] Step 5: Final Answer
\[ \frac{1}{2}|\mathbf{a} - \mathbf{b}| \]