Prove. Let x1, x2, ...... , xk be k pairwise orthogonal vectors in n-space. Then
1) ( x1 + x2 + ...... + xk ) · ( x1 + x2 + ...... + xk ) = x1 · x1 + x2 · x2 + ...... + xk · xk
i.e. the square of the diagonal of the k-dimensional rectangular parallelepiped formed by x1, x2, ...... , xk is equal to the sum of the squares of the edges.
Proof. The proof follows directly from the properties of the dot product, remembering that xi•xj = 0 when i j. For example, consider the case for three-dimensional space:
(x1 + x2 + x3)•(x1 + x2 + x3) = x1•(x1 + x2 + x3) + x2• (x1 + x2 + x3) + x3•(x1 + x2 + x3)
= x1•x1 + x1•x2 + x1•x3 + x2•x1 + x2•x2 + x2•x3 + x3•x1 + x3•x2 + x3•x3
= x1•x1 + x2•x2 + x3•x3