Question

If \(\overline{\mathrm{a}}\) and \(\overline{\mathrm{b}}\) are unit vectors such that \(|\overline{\mathrm{a}}+\overline{\mathrm{b}}|=\sqrt{3}\), then the angle between \(\bar{a}\) and \(\bar{b}\) is

• A. \(\frac{\pi}{6}\)
• B. \(\frac{\pi}{3}\)
• C. \(\frac{\pi}{4}\)
• D. \(\frac{\pi}{2}\)

Hint:

Square the magnitude of the sum and use the dot product of unit vectors.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
For any two vectors, \(|\bar{a}+\bar{b}|^{2}=|\bar{a}|^{2}+|\bar{b}|^{2}+2\,\bar{a}\cdot\bar{b}\).
Here \(|\bar{a}|=|\bar{b}|=1\), so \(\bar{a}\cdot\bar{b}=\cos\theta\).

Step 2: Calculation
Square the given condition: \(3=1+1+2\cos\theta\).
So \(2\cos\theta=1\) and \(\cos\theta=\dfrac{1}{2}\).

Step 3: Find the Angle
\(\cos\theta=\dfrac{1}{2}\) gives \(\theta=\dfrac{\pi}{3}\), which is option (B).

Final Answer:
The angle between the two unit vectors is \(\dfrac{\pi}{3}\). \[ \boxed{\text{(B) }\dfrac{\pi}{3}} \]