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(B) Let the cube have vertices at $(0,0,0)$ and $(a,a,a)$. The four diagonals are vectors along the directions: $\vec{d_1} = (a,a,a)$,$\vec{d_2} = (a,

Vedclass · Accepted Answer

$8/3$

Vedclass · Answer

(B) Let the cube have vertices at $(0,0,0)$ and $(a,a,a)$. The four diagonals are vectors along the directions: $\vec{d_1} = (a,a,a)$,$\vec{d_2} = (a,a,-a)$,$\vec{d_3} = (a,-a,a)$,and $\vec{d_4} = (-a,a,a)$.Let the direction of the given line be $\vec{l} = (x,y,z)$ where $x^2+y^2+z^2=1$.The cosine of the angle $	heta$ between the line and a diagonal $\vec{d}$ is given by $|\vec{l} \cdot \hat{d}|$.Thus,$\cos \alpha = \frac{|x+y+z|}{\sqrt{3}}$,$\cos \beta = \frac{|x+y-z|}{\sqrt{3}}$,$\cos \gamma = \frac{|x-y+z|}{\sqrt{3}}$,and $\cos \delta = \frac{|-x+y+z|}{\sqrt{3}}$.Summing the squares of the cosines:$\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma + \cos^2 \delta = \frac{1}{3} [(x+y+z)^2 + (x+y-z)^2 + (x-y+z)^2 + (-x+y+z)^2]$$= \frac{1}{3} [4(x^2+y^2+z^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 find the value of $\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma + \sin^2 \delta$.

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