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: 13

Answer

TRUE

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.
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