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 788: 23

Answer

$${\bf{u}} \cdot {\bf{v}} = - 4,\,\,\,\,{\text{and}}\,\,\,\,\theta \approx {104^ \circ }$$

Work Step by Step

$$\eqalign{ & {\bf{u}} = \left\langle {2,0, - 3} \right\rangle {\text{ and }}{\bf{v}} = \left\langle {1,4,2} \right\rangle \cr & {\text{find the dot product using the theorem 11}}{\text{.1 }}\left( {page\,\,783} \right) \cr & {\bf{u}} \cdot {\bf{v}} = \left\langle {2,0, - 3} \right\rangle \cdot \left\langle {1,4,2} \right\rangle = \left( 2 \right)\left( 1 \right) + \left( 0 \right)\left( 4 \right) + \left( { - 3} \right)\left( 2 \right) \cr & {\bf{u}} \cdot {\bf{v}} = 2 + 0 - 6 \cr & {\bf{u}} \cdot {\bf{v}} = - 4 \cr & {\text{find the magnitude of }}{\bf{u}}{\text{ and }}{\bf{v}}\,\,\left( {see\,\,page\,\,\,\,776} \right) \cr & \left| {\bf{u}} \right| = \left| {\left\langle {2,0, - 3} \right\rangle } \right| = \sqrt {{{\left( 2 \right)}^2} + {{\left( 0 \right)}^2} + {{\left( { - 3} \right)}^2}} = \sqrt {4 + 9} = \sqrt {13} \cr & \left| {\bf{v}} \right| = \left| {\left\langle {1,4,2} \right\rangle } \right| = \sqrt {{{\left( 1 \right)}^2} + {{\left( 4 \right)}^2} + {{\left( 2 \right)}^2}} = \sqrt {1 + 16 + 4} = \sqrt {21} \cr & {\text{find the angle between the vectores using }}\cos \theta = \frac{{{\bf{u}} \cdot {\bf{v}}}}{{\left| {\bf{u}} \right|\left| {\bf{v}} \right|}}{\text{ then}} \cr & \cos \theta = \frac{{{\bf{u}} \cdot {\bf{v}}}}{{\left| {\bf{u}} \right|\left| {\bf{v}} \right|}} = \frac{{ - 4}}{{\left( {\sqrt {13} } \right)\left( {\sqrt {21} } \right)}} \cr & {\text{solving for }}\theta \cr & \theta = {\cos ^{ - 1}}\left( {\frac{{ - 4}}{{\left( {\sqrt {13} } \right)\left( {\sqrt {21} } \right)}}} \right) \cr & {\text{simplify by using a calculator}} \cr & \theta \approx {104^ \circ } \cr} $$
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