Discussion
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
Answer: B
The angle is \(\frac{\pi}{6}\).
The projection of \(\vec{a}\) on \(\vec{b}\) is \(|\vec{a}|\cos\theta\):
\[ \begin{aligned} 12\cos\theta &= 6\sqrt{3} \\ \cos\theta &= \frac{\sqrt{3}}{2} \Rightarrow \theta = \frac{\pi}{6} \end{aligned} \]
\(\frac{\pi}{3}\) has cosine \(\frac{1}{2}\), which would need a projection of 6. A positive projection rules out the obtuse angles \(\frac{2\pi}{3}\) and \(\frac{3\pi}{4}\), and a right angle would give zero projection.
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