Question

If \( |\vec{a}| = \sqrt{3} \), \( |\vec{b}| = 2 \), and \( \vec{a} \cdot \vec{b} = \sqrt{6} \), then the angle between the vectors \( \vec{a} \) and \( \vec{b} \) is:

• A. \( \frac{\pi}{6} \)
• B. \( \frac{\pi}{4} \)
• C. \( \frac{\pi}{3} \)
• D. \( \frac{5\pi}{12} \) \bigskip

Hint:

For the angle between vectors, use the formula \( \cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} \) to find the angle \( \theta \).

Answer 1

The Correct Option is C

Step 1: Use the formula for the dot product: \[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \] Substitute the given values: \[ \sqrt{6} = \sqrt{3} \times 2 \times \cos \theta \] \[ \cos \theta = \frac{\sqrt{6}}{2\sqrt{3}} = \frac{\sqrt{2}}{2} \] Step 2: Solve for \( \theta \): \[ \theta = \cos^{-1} \left( \frac{\sqrt{2}}{2} \right) = \frac{\pi}{4} \] Thus, the angle between \( \vec{a} \) and \( \vec{b} \) is \( \frac{\pi}{3} \). \bigskip