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 $OADBFEGC$ be the cube with each side of length $a$ units.

The four diagonals are $OE,AF,BG$ and $CD$.
The direction cosines of the diagonal $OE$ 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 $AF,BG$ and $CD$ 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 $OE,AF,BG,CD$, 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
${\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma +{\cos }^2\delta$
$=\frac{1}{3}[(l+m+n{)}^2+(-l+m+n{)}^2+(l-m+n{)}^2+(I+m-n{)}^2]$
$=\frac{1}{3}\left[4\left(l^2+m^2+n^2\right)\right]=\frac{4}{3}$
$[$ as $I^2+m^2+n^2=1]$