A line makes angles $\alpha, \beta, \gamma, \delta$ with the four diagonals of a cube then $\cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma+\cos ^2 \delta=$
Answer & explanation
Correct answer: option 2
The direction ratios of the diagonal $\overrightarrow{OR}~ (1,1,1)$
Direction cosine are $\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)$
Similarly direction cosine of $\overrightarrow{AS}$ are $\left(\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}\right)$
$\overrightarrow{BP}$ are $\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}\right)$
$\overrightarrow{CQ}$ are $\left(\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)$
Let l, m, n be direction cosines of the line
$\cos \alpha=\frac{1+m+n}{\sqrt{3}}, \cos \beta=\frac{l-m-n}{\sqrt{3}}, \cos \gamma=\frac{l+m-n}{\sqrt{3}}, \cos \delta=\frac{l-m+n}{\sqrt{3}}$
$\cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma+\cos ^2 \delta=\frac{4\left(l^2+m^2+n^2\right)}{3}=\frac{4}{3}$ (since $l^2+m^2+n^2=1$)
Hence (2) is the correct answer.