Question

If `veca` and `vecb` are two vectors perpendicular to each other, prove that `(veca + vecb)^2 = (veca - vecb)^2`

Answer 1

`veca` and `vecb` are perpendicular to each other.

∴ `veca*vecb = vecb*veca` = 0    ...(i)

LHS = `(veca + vecb)^2`

= `(veca + vecb)*(veca + vecb)`

= `veca*(veca + vecb) + vecb(veca + vecb)`

= `veca*veca + veca*vecb + vecb*veca + vecb*vecb`

= `veca*veca + 0 + 0 + vecb*vecb`   ....[By (1)]

= `|veca|^2 + |vecb|^2`

RHS = `(veca - vecb)^2`

= `(veca - vecb)*(veca - vecb)`

= `veca*(veca - vecb) + vecb(veca - vecb)`

= `veca*veca - veca*vecb - vecb*veca + vecb*vecb`

= `veca*veca + vecb*vecb`     ....[By (i)]

= `|veca|^2 + |vecb|^2`

∴ LHS = RHS

Hence, `(veca + vecb)^2 = (veca - vecb)^2`