Discussion
Let \(|\vec{a}| = 2\) , \(|\vec{b}| = \sqrt{13 + 6\sqrt{3}}\) and \(|\vec{c}| = 3\) . If \(\vec{a} - \vec{b} + \vec{c} = 0\) , then the angle between \(\vec{a}\) and \(\vec{c}\) is
Answer: E
The angle between \(\vec{a}\) and \(\vec{c}\) is \(\dfrac{\pi}{6}\).
From \(\vec{a} - \vec{b} + \vec{c} = 0\), \(\vec{b} = \vec{a} + \vec{c}\). Square both sides:
\[ \begin{aligned} |\vec{b}|^{2} &= |\vec{a}|^{2} + |\vec{c}|^{2} + 2|\vec{a}||\vec{c}|\cos\theta \\ 13 + 6\sqrt{3} &= 4 + 9 + 12\cos\theta \\ \cos\theta &= \frac{\sqrt{3}}{2} \end{aligned} \]
So \(\theta = \frac{\pi}{6}\). \(\frac{\pi}{3}\) would need \(\cos\theta = \frac{1}{2}\), that is \(|\vec{b}|^{2} = 19\), and \(\frac{\pi}{2}\) would need \(|\vec{b}|^{2} = 13\).
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