Question

A lines makes angles $\alpha ,\beta ,\gamma ,\delta$ with the four diagonals cube. Then ${\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma +{\cos }^2\delta$ is

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

Answer 1

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

The directions of,

$OG:1,1,1$

$AE:-1,1,1$

$CD=1,1,-1$

$FB:-1,1,-1$

Let directions of line be $l,m,n$

$\therefore \cos \alpha =\frac{l+m+n}{\sqrt{3}}$ $(\alpha$ is angle made by $OG)$

$\therefore \cos \beta =\frac{-l+m+n}{\sqrt{3}}$ $(\beta$ is angle made by $AE)$

$\therefore \cos \gamma =\frac{l+m-n}{\sqrt{3}}$ $(\gamma$ is angle made by $CD)$

$\therefore \cos \delta =\frac{-l+m-n}{\sqrt{3}}$ $(\delta$ is angle made by $FB)$

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