Question

If `bar("a")` and `bar("b")` are two vectors perpendicular each other, prove that `(bar("a") + bar("b"))^2 = (bar("a") - bar("b"))^2`

Answer 1

`bar("a")` is perpendicular to `bar("b")`.

∴ `bar("a")*bar("b")` = 0

`(bar("a") + bar("b")) = (bar("a"))^2 + 2bar("a")*bar("b") + (bar("b"))^2`

= `(bar("a"))^2 + 2(0) + (bar("b"))^2`

= `(bar("a"))^2 + (bar("b"))^2`     .......(i)

`(bar("a") - bar("b"))^2 = (bar("a"))^2 - 2bar("a")*bar("b") + (bar("b"))^2`

= `(bar("a"))^2 + 2(0) + (bar("b"))^2`

= `(bar("a"))^2 + (bar("b"))^2`     .......(ii)

From (i) and (ii), we get

`(bar("a") + bar("b"))^2 = (bar("a") - bar("b"))^2`