Question

If a lines makes angles $\alpha ,\beta ,\gamma ,\delta$ with four digonals of a cube. Then ${\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma +{\cos }^2\delta$ will be :

• A. $4/3$
• B. $3/4$
• C. $1/4$
• D. None of these

Answer 1

Answer: (A)

$4/3$

$D{C}^{\prime }s$ of the four diagonals of a cube on putting $a=b=c$ 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 $l,m,n$ be the dc's of the line which is inclined at angles $\alpha ,\beta ,\gamma ,\delta$ resp to the four diagonals then
$\cos \alpha =\frac{1}{\sqrt{3}}\cdot l+m\cdot \frac{1}{\sqrt{3}}+n\cdot \frac{1}{\sqrt{3}}=\frac{l+m+n}{\sqrt{3}}$
Similarly $\cos \beta =\frac{-l+m-n}{\sqrt{3}},\cos \gamma =\frac{l-m+n}{\sqrt{3}},\cos \delta =\frac{l+m-n}{\sqrt{3}}$
$\therefore {\cos }^2\alpha ,{\cos }^2\beta +{\cos }^2\gamma +{\cos }^2\delta$
$=\frac{4(l^2+m^2+n^2)}{3}=\frac{4}{3}(\because l^2+m^2+n^2=1)$