If the vectors A and B are such that |A+B|=|A-B, then find the angle between vectors A and B?
A+B = P = (Ax+Bx)i +(Ay+By)j
A-B = Q = (Ax-Bx)i +(Ay-By)j
|P|^2 = (Ax+Bx)^2 + (Ay+By)^2
|Q|^2 = (Ax-Bx)^2 + (Ay-By)^2
those are equal?
then
Ax^2+2AxBx+Bx^2 + Ay^2+2AyBy+By^2
=
Ax^2-2AxBx+Bx^2 + Ay^2-2AyBy+By^2
--------------------------------- subtract = 0
0 = 4AxBx +4AyBy
SO
AxBx+AyBy = 0 !!!!
= dot product (scalar product) = |A||B| cos (theta)
cos (what? ) = 0 :)