Answer to Question #11937 in Vector Calculus for ahmer
To prove it for vectors
we can use scalar product (*,*) of vectors.
Let us mention the properies of
1) (a+b,c) = (a,c) + (b,c)
2) (a,b) = |a| * |b| *
cos(phi), where phi is the angle between vectors a and b
if a and b are perpendicular, then phi=90, and so cos(phi)=0.
(a,b) = |a| * |b| * 0 = 0.
Moreover, the angle between a and a
is 0, so cos(\phi) =1, whence
(a,a) = |a| * |a| * 1 =
Therefore if a and b are perpendicular, then
= (a,a) + (a,b) + (b,a) + (b,b)
= |a|^2 +
0 + 0 + |b|^2 =
= |a|^2 + |b|^2.
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