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Question

If a line makes angles α, β, γ, δ with four diagonals of a cube, then cos2 α + cos2 β + cos2 γ + cos2 δ is equal to

(a) 13

(b) 23

(c) 43

(d) 83

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Solution

c 43




Let a be the length of an edge of the cube and let one corner be at the origin as shown in the figure. Clearly, OP, AR, BS and CQ are the diagonals of the cube. The direction ratios of OP, AR, BS and CQ area-0, a-0, a-0, i.e. a, a, a0-a, a-0, a-0, i.e. -a, a, aa-0, 0-a, a-0, i.e. a, -a, aa-0, a-0, 0-a, i.e. a, a,-a Let the direction ratios of a line be proportional to l, m and n. Suppose this line makes angles α, β, γ and δ with OP, AR, BS and CQ, respectively.Now, α is the angle between OP and the line whose direction ratios are proportional to l, m and n. cos α = a.l+a.m+a.na2+a2+a2l2+m2+n2cos α = l + m + n3l2+m2+n2Since β is the angle between AR and the line with direction ratios proportional to l, m and n, we get cos β = -a.l+a.m+a.na2+a2+a2l2+m2+n2 cos β=-l+m+n3l2+m2+n2Similarly, cos γ = a.l-a.m+a.na2+a2+a2l2+m2+n2 cos γ=l - m + n3l2+m2+n2 cos δ = a.l+a.m-a.na2+a2+a2l2+m2+n2 cos δ=l+m - n3l2+m2+n2cos2α + cos2β + cos2γ+cos2δ =l+m+n23l2+m2+n2+-l+m+n23l2+m2+n2+I-m+n23l2+m2+n2+l+m - n23l2+m2+n2 =13l2+m2+n2l+m+n2+-l+m+n2+I-m+n2+l+m -n2 =13l2+m2+n24l2+m2+n2 =43



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