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(C) Let the cube have side length $a$. The coordinates of the vertices are $O(0,0,0), A(a,0,0), B(0,a,0), C(0,0,a), L(a,a,a), M(a,0,a), N(0,a,a), P(a,

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$4/3$

Vedclass · Answer

(C) Let the cube have side length $a$. The coordinates of the vertices are $O(0,0,0), A(a,0,0), B(0,a,0), C(0,0,a), L(a,a,a), M(a,0,a), N(0,a,a), P(a,a,0)$.The four diagonals are $OP, AN, BM, CL$. The direction vectors of these diagonals are:$\vec{d_1} = (a, a, a) \implies \hat{u_1} = \frac{1}{\sqrt{3}}(1, 1, 1)$$\vec{d_2} = (-a, a, a) \implies \hat{u_2} = \frac{1}{\sqrt{3}}(-1, 1, 1)$$\vec{d_3} = (a, -a, a) \implies \hat{u_3} = \frac{1}{\sqrt{3}}(1, -1, 1)$$\vec{d_4} = (a, a, -a) \implies \hat{u_4} = \frac{1}{\sqrt{3}}(1, 1, -1)$Let the direction cosines of the given line be $(l, m, n)$,where $l^2 + m^2 + n^2 = 1$.The cosine of the angle $	heta$ between the line and a diagonal with unit vector $\hat{u}$ is given by $|\cos 	heta| = |l u_x + m u_y + n u_z|$.Thus,$\cos^2 \alpha = \frac{1}{3}(l+m+n)^2$,$\cos^2 \beta = \frac{1}{3}(-l+m+n)^2$,$\cos^2 \gamma = \frac{1}{3}(l-m+n)^2$,$\cos^2 \delta = \frac{1}{3}(l+m-n)^2$.Summing these up:$\sum \cos^2 \alpha = \frac{1}{3} [(l+m+n)^2 + (-l+m+n)^2 + (l-m+n)^2 + (l+m-n)^2]$$= \frac{1}{3} [4(l^2+m^2+n^2)] = \frac{4}{3}(1) = \frac{4}{3}$.

If a line makes angles $\alpha, \beta, \gamma, \delta$ with the four diagonals of a cube,then find the value of $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma + \cos^2 \delta$.

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