Question

If \( \vec{a} \) and \( \vec{b} \) are two vectors such that \( |\vec{a}| = 1 \), \( |\vec{b}| = 2 \), and \( \vec{a} \cdot \vec{b} = \sqrt{3} \), then the angle between \( 2\vec{a} \) and \( -\vec{b} \) is:

• A. \( a^2 \)
• B. \( 2a^2 \)
• C. \( 3a^2 \)
• D. 0

Hint:

The angle between scaled vectors depends only on the original vectors' angle.

Answer 1

The Correct Option is C

Step 1: Find the angle between \( \vec{a} \) and \( \vec{b} \)
The dot product formula gives: \[ \vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta \implies \sqrt{3} = (1)(2)\cos\theta \implies \cos\theta = \frac{\sqrt{3}}{2}. \] Thus, \( \theta = \frac{\pi}{6} \).

Step 2: Angle between \( 2\vec{a} \) and \( -\vec{b} \)
Since \( 2\vec{a} \) and \( -\vec{b} \) involve a scalar multiplication, the angle becomes: \[ \pi - \frac{\pi}{6} = \frac{5\pi}{6}. \]
Step 3: Verify the options
The correct angle is \( \frac{5\pi}{6} \), matching option (C).