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Miscellaneous · Q32

Q.A line makes angles α,β,γ,δ\alpha,\beta,\gamma,\delta with the four diagonals of a cube. Prove that cos⁡2α+cos⁡2β+cos⁡2γ+cos⁡2δ=43\cos^2\alpha+\cos^2\beta+\cos^2\gamma+\cos^2\delta=\dfrac{4}{3}.

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✓ Free question

Take a cube of side 11 with one vertex at the origin and edges along the axes. Its four space diagonals have direction ratios (1,1,1)(1,1,1), (−1,1,1)(-1,1,1), (1,−1,1)(1,-1,1), (1,1,−1)(1,1,-1), each of magnitude 3\sqrt3.

Let the given line have direction cosines (l,m,n)(l,m,n), so l2+m2+n2=1l^2+m^2+n^2=1. Using cos⁡θ=∣l1l2+…∣\cos\theta=|l_1l_2+\dots| for the angle between two lines (Section 8), with the diagonals' unit direction vectors:

cos⁡α=l+m+n3,cos⁡β=−l+m+n3,cos⁡γ=l−m+n3,cos⁡δ=l+m−n3.\cos\alpha=\frac{l+m+n}{\sqrt3},\quad \cos\beta=\frac{-l+m+n}{\sqrt3},\quad \cos\gamma=\frac{l-m+n}{\sqrt3},\quad \cos\delta=\frac{l+m-n}{\sqrt3}.

Squaring and adding all four numerators:

(l+m+n)2+(−l+m+n)2+(l−m+n)2+(l+m−n)2.(l+m+n)^2+(-l+m+n)^2+(l-m+n)^2+(l+m-n)^2.

Expanding each square, every cross term (lmlm, mnmn, nlnl) appears twice with a ++ sign and twice with a −- sign across the four expansions, so all cross terms cancel, leaving

4(l2+m2+n2)=4(1)=4.4(l^2+m^2+n^2)=4(1)=4.

So cos⁡2α+cos⁡2β+cos⁡2γ+cos⁡2δ=43\cos^2\alpha+\cos^2\beta+\cos^2\gamma+\cos^2\delta=\dfrac{4}{3} (dividing the sum of squared numerators, 44, by the common denominator 33).

✓Final answer

cos⁡2α+cos⁡2β+cos⁡2γ+cos⁡2δ=43\cos^2\alpha+\cos^2\beta+\cos^2\gamma+\cos^2\delta=\dfrac{4}{3} (proved)

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