Question:
If \( |\vec{a}| = 12 \) and the projection of \( \vec{a} \) on \( \vec{b} \) is \( 6\sqrt{3} \), then the angle between \( \vec{a} \) and \( \vec{b} \) is
If \( |\vec{a}| = 12 \) and the projection of \( \vec{a} \) on \( \vec{b} \) is \( 6\sqrt{3} \), then the angle between \( \vec{a} \) and \( \vec{b} \) is
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Projection magnitude = \( |\vec{a}| \cos\theta \) (no need for \( \vec{b} \) magnitude).
Updated On: Apr 21, 2026
- \( \frac{\pi}{2} \)
- \( \frac{\pi}{6} \)
- \( \frac{\pi}{3} \)
- \( \frac{2\pi}{3} \)
- \( \frac{3\pi}{4} \)
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The Correct Option is B
Solution and Explanation
Concept:
Projection:
\[
\text{proj}_{\vec{b}} \vec{a} = |\vec{a}| \cos\theta
\]
Step 1: Apply formula.
\[ 12\cos\theta = 6\sqrt{3} \] \[ \cos\theta = \frac{\sqrt{3}}{2} \]
Step 2: Find angle.
\[ \theta = \frac{\pi}{6} \]
Step 1: Apply formula.
\[ 12\cos\theta = 6\sqrt{3} \] \[ \cos\theta = \frac{\sqrt{3}}{2} \]
Step 2: Find angle.
\[ \theta = \frac{\pi}{6} \]
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