Question

A line makes angles $\alpha, \beta, \gamma$ and $\delta$ with the diagonals of a cube. Then, $\cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma+\cos ^2 \delta$ is equal to

• A. $\frac{1}{3}$
• B. $\frac{3}{4}$
• C. $\frac{4}{3}$
• D. $\frac{8}{3}$

Answer 1

Answer: (C)

A cube is a rectangular parallelopiped having equal length, breadth and height. Let $O A D B F E G C$ be the cube with each side of length $a$ units.

The four diagonals are $O E, A F, B G$ and $C D$.
The direction cosines of the diagonal $O E$ which is the line joining two points $O$ and $E$ are
$ \frac{a-0}{\sqrt{a^2+a^2+a^2}}, \frac{a-0}{\sqrt{a^2+a^2+a^2}}, \frac{a-0}{\sqrt{a^2+a^2+a^2}} $
$\text { i.e., } \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}$
Similarly, the direction cosines of $A F, B G$ and $C D$ are $\frac{-1}{\sqrt{3}}$, $\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} ; \frac{1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}, \frac{1}{\sqrt{3}}$ and $\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}$, respectively.
Let $I, m, n$ be the direction cosines of the given line which makes angles $\alpha, \beta, \gamma, \delta$ with $O E, A F, B G, C D$, respectively. Then,
$ \cos \alpha=\frac{1}{\sqrt{3}}(l+m+n) ; \cos \beta=\frac{1}{\sqrt{3}}(-l+m+n) $
$ \cos \gamma=\frac{1}{\sqrt{3}}(l-m+n) ; \cos \delta=\frac{1}{\sqrt{3}}(l+m-n)$
Squaring and adding, we get
$ \cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma+\cos ^2 \delta $
$=\frac{1}{3}\left[(l+m+n)^2+(-l+m+n)^2+(l-m+n)^2\right.\left.+(I+m-n)^2\right]$
$=\frac{1}{3}\left[4\left(l^2+m^2+n^2\right)\right]=\frac{4}{3}$
$\left[\right.$ as $\left.I^2+m^2+n^2=1\right]$