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 - 11.2 Exercises - Page 764: 78

Answer

$$\eqalign{ & \left( {\bf{a}} \right)\left\langle {\frac{3}{5},0,\frac{4}{5}} \right\rangle \cr & \left( {\bf{b}} \right) - \left\langle {\frac{3}{5},0,\frac{4}{5}} \right\rangle \cr} $$

Work Step by Step

$$\eqalign{ & {\bf{v}} = \left\langle {6,0,8} \right\rangle \cr & {\text{Find }}\left\| {\bf{v}} \right\| \cr & \left\| {\bf{v}} \right\| = \sqrt {{{\left( 6 \right)}^2} + {{\left( 0 \right)}^2} + {{\left( 8 \right)}^2}} \cr & \left\| {\bf{v}} \right\| = \sqrt {100} \cr & \left\| {\bf{v}} \right\| = 10 \cr & \left( {\bf{a}} \right){\text{ A unit vector in the direction of }}{\bf{v}}{\text{ is}} \cr & \frac{1}{{\left\| {\bf{v}} \right\|}}{\bf{v}} = \frac{1}{{10}}\left\langle {6,0,8} \right\rangle \cr & \frac{1}{{\left\| {\bf{v}} \right\|}}{\bf{v}} = \left\langle {\frac{3}{5},0,\frac{4}{5}} \right\rangle \cr & \left( {\bf{b}} \right){\text{ A unit vector in the opposite direction of }}{\bf{v}}{\text{ is}} \cr & - \frac{1}{{\left\| {\bf{v}} \right\|}}{\bf{v}} = - \left\langle {\frac{3}{5},0,\frac{4}{5}} \right\rangle \cr} $$
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