If vector a,b,c are three mutually perpendicular vectors equally inclined to vector a + b + c at angle θ, then find the value of 36 cos^2 2θ.

Answer: 4

Let \(\vec s = \vec a + \vec b + \vec c\). Since the vectors are mutually perpendicular, \(\vec a\cdot\vec b = \vec b\cdot\vec c = \vec c\cdot\vec a = 0\), so

\[\vec a\cdot\vec s = |\vec a|^{2}\Rightarrow \cos\theta = \frac{|\vec a|}{|\vec s|}\]

Similarly \(\cos\theta = |\vec b|/|\vec s| = |\vec c|/|\vec s|\). Equal angles therefore force \(|\vec a| = |\vec b| = |\vec c| = k\).

Then \(|\vec s|^{2} = 3k^{2}\), so \(|\vec s| = \sqrt3\,k\) and

\[\cos\theta = \frac{1}{\sqrt3},\quad \cos^{2}\theta = \frac13\]

\[\cos2\theta = 2\cos^{2}\theta - 1 = -\frac13\]

\[36\cos^{2}2\theta = 36\times\frac19 = 4\]