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:
The cosine of the angle \( \theta \) between two vectors \( \vec{u} \) and \( \vec{v} \) is given by \( \cos \theta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}| |\vec{v}|} \).
Here, let \( \vec{u} = 2\vec{b} \) and \( \vec{v} = -\vec{a} \).

Step 2: Key Formula or Approach:
First, find the angle \( \phi \) between \( \vec{a} \) and \( \vec{b} \):
\[ \cos \phi = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{\sqrt{3}}{2 \cdot 1} = \frac{\sqrt{3}}{2} \implies \phi = \frac{\pi}{6} \]

Step 3: Detailed Explanation:
The dot product of \( 2\vec{b} \) and \( -\vec{a} \) is:
\[ (2\vec{b}) \cdot (-\vec{a}) = -2 (\vec{a} \cdot \vec{b}) = -2 (\sqrt{3}) = -2\sqrt{3} \]
The magnitudes are:
\[ |2\vec{b}| = 2|\vec{b}| = 2 \quad \text{and} \quad |-\vec{a}| = |\vec{a}| = 2 \]
Using the angle formula:
\[ \cos \theta = \frac{-2\sqrt{3}}{2 \cdot 2} = \frac{-2\sqrt{3}}{4} = -\frac{\sqrt{3}}{2} \]
Since \( \cos \theta = -\frac{\sqrt{3}}{2} \), the angle \( \theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6} \).

Step 4: Final Answer:
The angle between \( 2\vec{b} \) and \( -\vec{a} \) is \( \frac{5\pi}{6} \).