A line makes angles α, β, γ, δ with the diagonals of a cube. Prove that cos^2α + cos^2β + cos^2γ + cos^2δ = 4/3.

Question

A line makes angles α, β, γ, δ with the diagonals of a cube. Prove that cos²α + cos²β + cos²γ + cos²δ = \(\frac 43\).

Answer 1

Let OALBPNCN be a cube with each side of length a.

Now the coordinates are as points O(0,0,0), A (a, 0, 0), B(0, a, 0), C(0, 0, a), L(a, a, 0) , N(0, a, a), M(a, 0, a), P (a, a, a). Obviously, the four diagonals of this cube are OP, AN, BN and CL.

Direction cosine of the diagonal OP is

\(\frac{a - 0}{\sqrt{a^2 + a^2 + a^2}}, \frac{a - 0}{\sqrt{a^2 + a^2 + a^2}}, \frac{a - 0}{\sqrt{a^2 + a^2 + a^2}}\)

That means \(\frac 1{\sqrt 3},\frac 1{\sqrt 3},\frac 1{\sqrt 3}\). 

Direction cosine of the diagonal AN is

\(\frac{a - 0}{\sqrt{a^2 + a^2 + a^2}}, \frac{a - 0}{\sqrt{a^2 + a^2 + a^2}}, \frac{ 0-a}{\sqrt{a^2 + a^2 + a^2}}\)

That means  \(\frac 1{\sqrt 3},\frac 1{\sqrt 3},\frac {-1}{\sqrt 3}\).

Suppose the direction cosine of the given line is l, m, n which make angles α, β, γ, δ with the diagonals OP, AM, BN, CL respectively, then

\(\cos \alpha = \left(l.\frac 1{\sqrt 3} + m.\frac 1{\sqrt 3} + n.\frac 1{\sqrt 3}\right) = \frac{(l + m+n)}{\sqrt 3}\)

\(\cos \beta = \left(l.(\frac {-1}{\sqrt 3}) + m.\frac 1{\sqrt 3} + n.\frac 1{\sqrt 3}\right) = \frac{(-l + m+n)}{\sqrt 3}\)

\(\cos \gamma = \left(l.\frac {1}{\sqrt 3} + m.(\frac {-1}{\sqrt 3}) + n.\frac 1{\sqrt 3}\right) = \frac{(l - m+n)}{\sqrt 3}\)

\(\cos \delta = \left(l.\frac {1}{\sqrt 3} + m.\frac {1}{\sqrt 3} + n.(\frac {-1}{\sqrt 3})\right) = \frac{(l + m-n)}{\sqrt 3}\)

On squaring and adding, we get

cos²α + cos²β + cos²γ + cos²δ

\(=\frac 13[(l + m+ n)^2 + (- l + m+n)^2 + (l - m + n)^2 + (l + m - n)^2]\)

\(= \frac 43\)