Question

A line makes angles $\alpha ,\beta ,\gamma ,\delta$ with the four diagonals of a cube. Then $co{s}^2\alpha +co{s}^2\beta +co{s}^2\gamma +co{s}^2\delta$ is

• A. $\frac{4}{3}$
• B. $\frac{2}{3}$
• C. $3$
• D. $0$

Answer 1

The correct option is A $\frac{4}{3}$

DC’s of diagonals OP, BQ, CS and AR are
$\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}};\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}};\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}};-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$
Let dc’s of a line be l, m, n
$\therefore cos\alpha =\frac{l+m+n}{\sqrt{3}}cos\beta =\frac{l-m+n}{\sqrt{3}}cos\gamma =\frac{l+m-n}{\sqrt{3}}cos\delta =\frac{-l+m+n}{\sqrt{3}}$

$\therefore cos2\alpha +cos2\beta +cos2\gamma =\frac{(l+m+n{)}^2+(l-m+n{)}^2+(l+m-n{)}^2+(-l+m+n{)}^2}{3}=\frac{4}{3}$