Question

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

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

Answer 1

The correct option is C $\frac{8}{3}$

Diagonals are $OE=(1,1,1)$

$\overrightarrow{GB}=(1,1,0)-(0,0,1)=(1,1,-1)$

$\overrightarrow{FC}=(1,-1,1)$

$\overrightarrow{AD}=(-1,1,1)$

Let the line have direction ratio's as $l,m,n$

Angle between $OE$ and $L$

$\cos \alpha =\frac{l+m+n}{\sqrt{l^2+m^2+n^2}\times 2\sqrt{3}}$

$\cos \beta =\frac{l+m-n}{\sqrt{l^2+m^2+n^2}\times \sqrt{3}}$

$\cos \gamma =\frac{l-m+n}{\sqrt{l^2+m^2+n^2}\times \sqrt{3}}$

$\cos \delta =\frac{-l+m+n}{\sqrt{l^2+m^2+n^2}\times \sqrt{3}}$

${\sin }^2\alpha +{\sin }^2\beta +{\sin }^2\gamma +{\sin }^2\delta =4-({\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma +{\cos }^2\delta )$

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

$=\frac{8}{3}$