Answer to Question #11937 in Vector Calculus for ahmer

Question

if a and b are perpendicular then prove that [a2+b2]=[a]2+[b]2

Answer 1

This result is a varaint of Pythagorean theorem.

To prove it for vectors
we can use scalar product (*,*) of vectors.
Let us mention the properies of
this product:

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

In particular,
if a and b are perpendicular, then phi=90, and so cos(phi)=0.
Therefore


(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 =
|a|^2.


Therefore if a and b are perpendicular, then

|a+b|^2 =
(a+b, a+b)
= (a,a) + (a,b) + (b,a) + (b,b)
= |a|^2 +
0 + 0 + |b|^2 =
= |a|^2 + |b|^2.