Prove that alpha^2 + beta^2 + gamma ^2 = 1?

Question

The direction cosines of a vector are the cosines of the angles it makes with the coordinate axes. The cosine of the angles between the vector and the x, y, and z axes are usually called in turn alpha, beta, and gamma. Prove that alpha^2 + beta^2 + gamma ^2 = 1, using either geometry or vector algebra.
(more interested in how to prove it with vector algebra, but geometry is fine too.)

Answer 1

Let the vector be v = (x, y, z). Recall that the dot product obeys the rule
a·b = ||a|| ||b|| cos θ
where θ is the angle between a and b.
Hence (v·i)^2 = ||v||^2 α^2 and similarly for β and γ.
v·i = x, v·j = y, and v·k = z.
So ||v||^2 (α^2 + β^2 + γ^2)
= (v·i)^2 + (v·j)^2 + (v·k)^2
= (x^2 + y^2 + z^2)
= ||v||^2
So α^2 + β^2 + γ^2 = 1.

Answer 2

For any vector R of magnitude r with scalar components (r1, r2, r3) making angles alpha, beta, gamma with the co-ordinate axes:
r^2 = r1^2 + r2^2 + r3^2
r^2 = R.R
= (r cos(alpha), r cos(beta), r cos(gamma))^2
= r^2 cos^2(alpha) + r^2 cos^2(beta) + r^2 cos^2(gamma)
= r^2 (cos^2(alpha) + cos^2(beta) + cos^2(gamma))
Hence:
cos^2(alpha) + cos^2(beta) + cos^2(gamma)) = 1.

Answer 3

I would submit all the permutations to a group theory analysis, and eliminate other answers.