Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 12 - Vectors and the Geometry of Space - Review - True-False Quiz - Page 882: 13



Work Step by Step

Cross product is also known as vector product which is perpendicular to two vectors , therefore, it must also be perpendicular to either vector , that is , $u$ is orthogonal to $u\times v$. 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.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.