Vectorz pls helppp?

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

ok thanx SS

all others dont answer therez no use SS is gonna get the 10pts

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.

Answer 2

squaring both sides,
a^2 + b^2 + 2mod(a)mod(b) = a^2 + b^2 - 2mod(a)mod(b)
4mod(a)mod(b)=0
mod(a)mod(b)=0
now since dot product of 2 vectors a & b =0 therefore angle b/w them is 90deg
cosx=0 i.e x=90deg.