## Calculus: Early Transcendentals 8th Edition

Cross product is also known as vector product which is perpendicular to two vectors; therefore, it must also be perpendicular to either vector Consider $u=( u_{1},u_{2},u_{3})$ and $v=( v_{1},v_{2},v_{3})$ then $(u\times v).u=( u_{2}v_{3}-u_{3}v_{2},- (u_{1}v_{3}-u_{3}v_{1}), u_{1}v_{2}-u_{2}v_{1})( u_{1},u_{2},u_{3})$ $=u_{1}( u_{2}v_{3}-u_{3}v_{2})- u_{2}(u_{1}v_{3}-u_{3}v_{1})+u_{3}(u_{1}v_{2}-u_{2}v_{1})$ $=u_{1}u_{2}v_{3}-u_{1}u_{3}v_{2}- u_{2}u_{1}v_{3}+u_{2}u_{3}v_{1}+u_{3}u_{1}v_{2}-u_{3}u_{2}v_{1}$ Therefore, $(u\times v).u=0$ Hence, the statement is true.