Question

If a line makes angles $\alpha, \beta, \gamma, \delta$ with four diagonals of a cube then value of
$sin^{2} \alpha+sin^{2}\beta +sin^{2}\gamma +sin^{2} \delta$ equals

• A. $2$
• B. $\frac{4}{3}$
• C. $\frac{8}{3}$
• D. $1$

Answer 1

Answer: (C)

Four diagonals are $OP, AA', BB', CC' $
dc’s of OP are $<\,\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} >$
dc’s of AA' are $<\,-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}>$
dc’s of BB' are $<\,\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}>$
dc’s of C'C are $<\, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}>$
Let $l, m, n$ be the dc ’s of line which makes the angle $\alpha, \beta, \delta $ with the four diagonals.

$\therefore cos\,\alpha = \frac{l+m+n}{\sqrt{3}}$,
$cos\,\beta = \frac{-l+m+m}{\sqrt{3}}$,
$cos\,\gamma=\frac{l-m+n}{\sqrt{3}}$,
$cos\,\delta =\frac{l+m-n}{\sqrt{3}}$
$\therefore cos^{2}\alpha+cos^{2}\beta+cos^{2}\gamma+cos^{2}\delta$
$=\left(\frac{1}{\sqrt{3}}\right)^{2} \left[1+1+1+1\right]=\frac{4}{3}$
$\therefore sin^{2}\alpha+sin^{2}\beta+sin^{2}\gamma+sin^{2}\delta=4-\frac{4}{3}=\frac{8}{3}$
Short Cut Method :
In such type of problem we use the well known feet
$cos^{2} \alpha+cos^{2}\beta +cos^{2}\gamma+cos^{2}\delta=\frac{4}{3}$
where $\alpha, \beta, \gamma, \delta$ are angles made by a line with four diagonals of a cube
$\therefore (1-sin^{2}\alpha) + (1-sin^{2}\beta) + (1-sin^{2} \gamma + (1-sin^{2}\delta) = \frac{4}{3}$
$\Rightarrow sin^{2}\alpha +sin^{2} \beta+sin^{2} \gamma+sin^{2} \delta=4-\frac{4}{3} =\frac{8}{3}$