Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 11 - Vectors and Vector-Valued Functions - 11.3 Dot Products - 11.3 Exercises - Page 789: 32

Answer

$${\text{pro}}{{\text{j}}_{\bf{v}}}{\bf{u}} = \left\langle { - 4,1,3} \right\rangle {\text{ and sca}}{{\text{l}}_{\bf{v}}}{\bf{u}} = - \sqrt {26} $$

Work Step by Step

$$\eqalign{ & {\bf{u}} = \left\langle {13,0,26} \right\rangle {\text{ and }}{\bf{v}} = \left\langle {4, - 1, - 3} \right\rangle \cr & {\text{The orthogonal projection of }}{\bf{u}}{\text{ onto }}{\bf{v}}{\text{, denoted pro}}{{\text{j}}_{\bf{v}}}{\bf{u}}, \cr & {\text{,where }}{\bf{v}} \ne 0,{\text{is}} \cr & {\text{pro}}{{\text{j}}_{\bf{v}}}{\bf{u}} = {\text{sca}}{{\text{l}}_{\bf{v}}}{\bf{u}}\left( {\frac{{\bf{v}}}{{\left| {\bf{v}} \right|}}} \right) = \left( {\frac{{{\bf{u}} \cdot {\bf{v}}}}{{{\bf{v}} \cdot {\bf{v}}}}} \right){\bf{v}} \cr & {\text{The scalar component of }}{\bf{u}}{\text{ in the direction of }}{\bf{v}}{\text{ is}} \cr & {\text{sca}}{{\text{l}}_{\bf{v}}}{\bf{u}} = \frac{{{\bf{u}} \cdot {\bf{v}}}}{{\left| {\bf{v}} \right|}} = \frac{{\left\langle {13,0,26} \right\rangle \cdot \left\langle {4, - 1, - 3} \right\rangle }}{{\left| {\left\langle {4, - 1, - 3} \right\rangle } \right|}} = - \frac{{26}}{{\sqrt {26} }} \cr & {\text{sca}}{{\text{l}}_{\bf{v}}}{\bf{u}} = - \sqrt {26} \cr & {\text{Calculating pro}}{{\text{j}}_{\bf{v}}}{\bf{u}} \cr & {\text{pro}}{{\text{j}}_{\bf{v}}}{\bf{u}} = {\text{sca}}{{\text{l}}_{\bf{v}}}{\bf{u}}\left( {\frac{{\bf{v}}}{{\left| {\bf{v}} \right|}}} \right) = - \sqrt {26} \left( {\frac{{\left\langle {4, - 1, - 3} \right\rangle }}{{\sqrt {26} }}} \right) \cr & {\text{pro}}{{\text{j}}_{\bf{v}}}{\bf{u}} = - \left\langle {4, - 1, - 3} \right\rangle \cr & {\text{pro}}{{\text{j}}_{\bf{v}}}{\bf{u}} = \left\langle { - 4,1,3} \right\rangle \cr} $$
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