Question #11937

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

Expert's answer

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.

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.

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