Question

Let \(\vec{a}, \vec{b}\) and \(\vec{c}\) be three vectors such that \(|\vec{a}|=|\vec{b}|=\frac{|\vec{c}|}{2}=1\). If \(\vec{a} \times(\vec{a} \times \vec{c})+\vec{b}=\overrightarrow{0}\), then the acute angle between \(\vec{a}\) and \(\vec{c}\) is :

• A. \(\frac{\pi}{6}\)
• B. \(\frac{\pi}{4}\)
• C. \(\frac{\pi}{3}\)
• D. \(\frac{5\pi}{12}\)

Hint:

Expand the vector triple product \(\vec a\times(\vec a\times\vec c)=(\vec a\cdot\vec c)\vec a-(\vec a\cdot\vec a)\vec c\) and take magnitudes.

Answer 1

The Correct Option is A

Using \(\vec a\times(\vec a\times\vec c)=(\vec a\cdot\vec c)\vec a-(\vec a\cdot\vec a)\vec c\) with \(|\vec a|=1\):
\((\vec a\cdot\vec c)\vec a-\vec c+\vec b=\vec 0\Rightarrow \vec b=\vec c-(\vec a\cdot\vec c)\vec a\).
The right side is the component of \(\vec c\) perpendicular to \(\vec a\), whose magnitude is \(|\vec c|\sin\theta=2\sin\theta\), where \(\theta\) is the angle between \(\vec a\) and \(\vec c\).
Since \(|\vec b|=1\), \(2\sin\theta=1\Rightarrow\sin\theta=\frac12\).
The acute angle is \(\theta=\frac{\pi}{6}\).
Answer: (1) \(\frac{\pi}{6}\)