Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 11 - Vectors and the Geometry of Space - 11.3 Exercises - Page 773: 42

Answer

$$\eqalign{ & \left( {\bf{a}} \right){{\bf{w}}_1} = \frac{{33}}{{13}}{\bf{i}} + \frac{{22}}{{13}}{\bf{k}} \cr & \left( {\bf{b}} \right){{\bf{w}}_2} = - \frac{{20}}{{13}}{\bf{i}} + \frac{{30}}{{13}}{\bf{j}} \cr} $$

Work Step by Step

$$\eqalign{ & {\text{Let the vectors be }}{\bf{u}} = {\bf{i}} + 4{\bf{k}},{\text{ }}{\bf{v}} = 3{\bf{i}} + 2{\bf{k}} \cr & \left( {\bf{a}} \right){\text{ Let }}{{\bf{w}}_1} = {\text{pro}}{{\text{j}}_{\bf{v}}}{\bf{u}} = \left( {\frac{{{\bf{u}} \cdot {\bf{v}}}}{{{{\left\| {\bf{v}} \right\|}^2}}}} \right){\bf{v}} \cr & {{\bf{w}}_1} = \left( {\frac{{\left( {{\bf{i}} + 4{\bf{k}}} \right) \cdot \left( {3{\bf{i}} + 2{\bf{k}}} \right)}}{{{{\left\| {\left( {3{\bf{i}} + 2{\bf{k}}} \right)} \right\|}^2}}}} \right)\left( {3{\bf{i}} + 2{\bf{k}}} \right) \cr & {{\bf{w}}_1} = \left( {\frac{{3 + 8}}{{9 + 4}}} \right)\left( {3{\bf{i}} + 2{\bf{k}}} \right) \cr & {{\bf{w}}_1} = \frac{{11}}{{13}}\left( {3{\bf{i}} + 2{\bf{k}}} \right) \cr & {{\bf{w}}_1} = \frac{{33}}{{13}}{\bf{i}} + \frac{{22}}{{13}}{\bf{k}} \cr & \cr & \left( {\bf{b}} \right){\text{From the Definitions of Projection and Vector Components}} \cr & {{\bf{w}}_2} = {\bf{u}} - {{\bf{w}}_1}{\text{ is called the }}{\bf{vector}}{\text{ }}{\bf{component}}{\text{ }}{\bf{of}}{\text{ }}{\bf{u}}{\text{ }}{\bf{orthogonal}} \cr & {\bf{to}}{\text{ }}{\bf{v}}. \cr & {{\bf{w}}_2} = \left( {{\bf{i}} + 4{\bf{k}}} \right) - \left( {\frac{{33}}{{13}}{\bf{i}} + \frac{{22}}{{13}}{\bf{k}}} \right) \cr & {{\bf{w}}_2} = - \frac{{20}}{{13}}{\bf{i}} + \frac{{30}}{{13}}{\bf{j}} \cr} $$
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