If |→a+→b|=|→a−→b|,then the vectors →a and →bare orthogonal
True
Since, |→a+→b|=|→a−→b|⇒ |→a+→b|2=|→a−→b|2⇒ 2|→a||→b|=−2|→a||→b|⇒ 4|→a||→b|=0⇒ |→a||→b|=0Hence, →a and →b are orthogonal. [∵→a.→b=|→a||→b|cos900=0]