Question

Let $\theta$ be the angle between the unit vectors $\hat{a}$ and $\hat{b}$. If $|\hat{a} - \hat{b}| = \frac{\sqrt{3}}{2}$, then the value of $\cos \theta$ is

• A. $\frac{3}{8}$
• B. $\frac{1}{2}$
• C. $\frac{5}{8}$
• D. $\frac{3}{4}$
• E. $\frac{7}{8}$

Hint:

For any unit vectors $\hat{a}, \hat{b}$, the identity \( |\hat{a} - \hat{b}| = 2 \sin(\theta/2) \) is very useful. Here \( 2 \sin(\theta/2) = \sqrt{3}/2 \implies \sin(\theta/2) = \sqrt{3}/4 \). Then use \( \cos \theta = 1 - 2 \sin^2(\theta/2) \).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
For unit vectors, \( |\hat{a}| = |\hat{b}| = 1 \). The magnitude of the difference of two vectors is given by \( |\hat{a} - \hat{b}|^2 = |\hat{a}|^2 + |\hat{b}|^2 - 2 \hat{a} \cdot \hat{b} \), where \( \hat{a} \cdot \hat{b} = |\hat{a}||\hat{b}| \cos \theta \).

Step 2: Detailed Explanation:
Given \( |\hat{a} - \hat{b}| = \frac{\sqrt{3}}{2} \).
Square both sides:
\[ |\hat{a} - \hat{b}|^2 = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4} \]
Expand the LHS:
\[ |\hat{a}|^2 + |\hat{b}|^2 - 2 \hat{a} \cdot \hat{b} = \frac{3}{4} \]
Since \( \hat{a} \) and \( \hat{b} \) are unit vectors:
\[ 1 + 1 - 2(1)(1) \cos \theta = \frac{3}{4} \]
\[ 2 - 2 \cos \theta = \frac{3}{4} \]
\[ 2 \cos \theta = 2 - \frac{3}{4} \]
\[ 2 \cos \theta = \frac{8 - 3}{4} = \frac{5}{4} \]
\[ \cos \theta = \frac{5}{8} \]

Step 3: Final Answer:
The value of \( \cos \theta \) is \( 5/8 \).