Question

If a line makes angles $\alpha ,\beta ,\gamma ,\delta$ with four diagonals of 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

Answer: (C)

Four diagonals are $OP,A{A}^{\prime },B{B}^{\prime },C{C}^{\prime }$
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.

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