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(C) Let the side of the cube be $a$. The four diagonals of the cube can be represented by vectors along the directions $(1, 1, 1)$,$(1, 1, -1)$,$(-1,

Vedclass · Accepted Answer

$\frac{8}{3}$

Vedclass · Answer

(C) Let the side of the cube be $a$. The four diagonals of the cube can be represented by vectors along the directions $(1, 1, 1)$,$(1, 1, -1)$,$(-1, 1, 1)$,and $(1, -1, 1)$.Let the direction cosines of the line be $(l, m, n)$,where $l^2 + m^2 + n^2 = 1$.The cosines of the angles $\alpha, \beta, \gamma, \delta$ between the line and the diagonals are given by:$\cos \alpha = \frac{|l + m + n|}{\sqrt{3}}$,$\cos \beta = \frac{|l + m - n|}{\sqrt{3}}$,$\cos \gamma = \frac{|-l + m + n|}{\sqrt{3}}$,$\cos \delta = \frac{|l - m + n|}{\sqrt{3}}$.Squaring these,we get:${\cos ^2}\alpha = \frac{(l + m + n)^2}{3}$,${\cos ^2}\beta = \frac{(l + m - n)^2}{3}$,${\cos ^2}\gamma = \frac{(-l + m + n)^2}{3}$,${\cos ^2}\delta = \frac{(l - m + n)^2}{3}$.Summing these:$\sum \cos^2 \alpha = \frac{1}{3} [ (l^2+m^2+n^2 + 2lm + 2mn + 2nl) + (l^2+m^2+n^2 + 2lm - 2mn - 2nl) + (l^2+m^2+n^2 - 2lm + 2mn - 2nl) + (l^2+m^2+n^2 - 2lm - 2mn + 2nl) ]$$= \frac{1}{3} [ 4(l^2+m^2+n^2) ] = \frac{4}{3} (1) = \frac{4}{3}$.Since ${\sin ^2}	heta = 1 - {\cos ^2}	heta$,we have:$\sum \sin^2 \alpha = 4 - \sum \cos^2 \alpha = 4 - \frac{4}{3} = \frac{8}{3}$.

If a line makes angles $\alpha, \beta, \gamma, \delta$ with the four diagonals of a cube,then the value of ${\sin ^2}\alpha + {\sin ^2}\beta + {\sin ^2}\gamma + {\sin ^2}\delta$ is

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