iii Vectors iii?

Question

_ _ _ _
Given : | a + b | = | a - b |

( it is actually mod of vector a + vector b = mod of vector a - vector b )
To prove : -

both the vectors are perpendicular

Answer 1

As you didn't specify, I'm going to assume we're working with 2-dimensional vectors.

Let (a1, a2) and (b1, b2) denote the components of a and b, respectively. Then since | a + b | = | a - b |, we know that | a + b |² = | a - b |², which we can write as

(a1 + b1)² + (a2 + b2)² = (a1 - b1)² + (a2 - b2)².

Expanding both sides yields

a1² + 2a1b1 + b1 ² + a2² + 2a2b2 + b2²
= a1² - 2a1b1 + b1 ² + a2² - 2a2b2 + b2²,

and eliminating the squares from both sides and gathering terms on the LHS yields

4(a1b1 + a2b2) = 0.

Thus

a1b1 + a2b2 = 0.

This is equivalent to saying that a and b are perpendicular (their inner product/scalar product is zero). The case for more than two dimensions is similar.