Question

Let \(\vec{a}, \vec{b}, \vec{c}\) be three mutually perpendicular vectors of the same magnitude and equally inclined at an angle \(\theta\), with the vector \(\vec{a} + \vec{b} + \vec{c}\). Then \(36\cos^2 2\theta\) is equal to ___________.

Hint:

Find \(\cos\theta\) from \(\vec{a}\cdot(\vec{a}+\vec{b}+\vec{c})\), then use \(\cos 2\theta = 2\cos^2\theta - 1\).

Answer 1

Correct Answer: 4

Let \(|\vec{a}| = |\vec{b}| = |\vec{c}| = k\). Since they are mutually perpendicular, \(|\vec{a}+\vec{b}+\vec{c}|^2 = 3k^2\), so \(|\vec{a}+\vec{b}+\vec{c}| = \sqrt{3}k\).
\(\cos\theta = \frac{\vec{a}\cdot(\vec{a}+\vec{b}+\vec{c})}{|\vec{a}||\vec{a}+\vec{b}+\vec{c}|} = \frac{k^2}{k\cdot\sqrt{3}k} = \frac{1}{\sqrt{3}}\).
\(\cos 2\theta = 2\cos^2\theta - 1 = \frac{2}{3} - 1 = -\frac{1}{3}\).
\(36\cos^2 2\theta = 36\cdot\frac{1}{9} = 4\).
Answer: 4