how do you prove that cosα^2 + cosΒ^2 + cosγ^2 = 1?

Question

how do you prove that cosα^2 + cosΒ^2 + cosγ^2 = 1?
This is direction cosines.

Answer 1

The direction cosines for point (x,y,z) in space are defined such that

x/r = cos alpha
y/r = cos beta
z/r = cos gamma

where r is the distance from the point to the origin.

According to the Pythagorean Theorem, the distance r is equal to
r^2 = x^2 + y^2 + z^2

Divide by r^2, you find
1 = (x/r)^2 + (y/r)^2 + (z/r)^2

The three terms on the right are precisely the squares of the direction cosines. This proves your statement.

Answer 2

Let the vector AB = (ip +jq +kr) is making angles a, b ,c with x-Axis, y-ax ix and with z-ax ix
Modulus of vector AB=sq rt(p^2+q^2+r^2) then direction cosines are
cos a = {p/sq rt(p^2+q^2+r^2)}.............i
cos b = {q/sq rt(p^2+q^2+r^2)}..............ii
cos c ={r/sq rt(p^2+q^2+r^2)}...............iii
Squaring and adding we get
cosα^2 + cosΒ^2 + cosγ^2
= (p^2+q^2+r^2)}/(p^2+q^2+r^2)}
=1

Answer 3

Here you are:
http://mathworld.wolfram.com/DirectionCosine.html