Question

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

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

Hint:

Remember that \( \vec{u} \cdot \vec{v} = |\vec{u}||\vec{v}|\cos\theta \). Always check the signs of the magnitudes and the dot product carefully when dealing with negative vectors.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Let \( \theta \) be the angle between \( \vec{a} \) and \( \vec{b} \). Using the definition of dot product, \( \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\theta \).

Step 2: Detailed Explanation:
Given: \( \sqrt{3} = 2 \cdot 1 \cdot \cos\theta \implies \cos\theta = \frac{\sqrt{3}}{2} \implies \theta = \frac{\pi}{6} \).
Let \( \vec{u} = 2\vec{b} \) and \( \vec{v} = -\vec{a} \).
The cosine of the angle \( \phi \) between \( \vec{u} \) and \( \vec{v} \) is:
\( \cos\phi = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}| |\vec{v}|} = \frac{(2\vec{b}) \cdot (-\vec{a})}{|2\vec{b}| |-\vec{a}|} = \frac{-2(\vec{a} \cdot \vec{b})}{2|\vec{b}| \cdot |\vec{a}|} = \frac{-2\sqrt{3}}{2 \cdot 1 \cdot 2} = -\frac{\sqrt{3}}{2} \).
Since \( \cos\phi = -\frac{\sqrt{3}}{2} \), the angle \( \phi = \pi - \frac{\pi}{6} = \frac{5\pi}{6} \).

Step 3: Final Answer:
The angle is \( \frac{5\pi}{6} \).