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Prove. Let x_{1}, x_{2}, ...... , x_{k} be k pairwise orthogonal vectors in n-space. Then

1) ( x_{1 }+ x_{2} + ...... + x_{k} ) · ( x_{1 }+ x_{2} + ...... + x_{k} ) = x_{1} ·_{ }x_{1} + x_{2} · x_{2} + ...... + x_{k} · x_{k}

i.e. the square of the diagonal of the k-dimensional rectangular parallelepiped formed by x_{1}, x_{2},
...... , x_{k} 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 x_{i}•x_{j}
= 0 when i
j. For example, consider the case for three-dimensional space:

(x_{1} + x_{2} + x_{3})•(x_{1} + x_{2} + x_{3}) = x_{1}•(x_{1} + x_{2} + x_{3}) + x_{2}• (x_{1} + x_{2} + x_{3}) + x_{3}•(x_{1} + x_{2} +
x_{3})

= x_{1}•x_{1} + x_{1}•x_{2} + x_{1}•x_{3} + x_{2}•x_{1} + x_{2}•x_{2} + x_{2}•x_{3} + x_{3}•x_{1} + x_{3}•x_{2} + x_{3}•x_{3}

= x_{1}•x_{1} + x_{2}•x_{2} + x_{3}•x_{3}

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