# Vector ...proof the pythagoras theorem?

the sides of triangle ABC are represented by th Vectors AB=a BC=b CA= c

Prove Pythagoras Theorem IaI^2 + IbI^2 = IcI^2, if angle ABC 90

### 2 Answers

- 1 decade agoFavourite answer
We'll use * to denote the dot product ( http://mathworld.wolfram.com/DotProduct.html ).

By vector addition, c = a + b.

Now, the square of the magnitude of c is

| c |^2 = c * c

= ( a + b ) * ( a + b )

= a * a + b * b + 2 a * b

= | a |^2 + | b |^2,

since a * b = 0 because a is perpendicular to b (the angle between them is 90 degrees). If the angle theta between them is not 90 degrees, you get

| c |^2 = | a |^2 + | b |^2 + 2 | a | | b | cos(theta),

the law of cosines.

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- 4 years ago
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