Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 8 - Application of Trigonometry - 8.4 Algebraically Defined Vectors and the Dot Product - 8.4 Exercises - Page 790: 33

Answer

$$\left. a \right)\left\langle { - 2,4} \right\rangle ,\,\,\,\,\,\left. b \right)\left\langle {7,4} \right\rangle ,\,\,\,\,\,\left. c \right)\left\langle {6, - 6} \right\rangle $$

Work Step by Step

$$\eqalign{ & {\text{Let }}{\bf{u}} = \left\langle { - 1,2} \right\rangle {\text{ and }}{\bf{v}} = \left\langle {3,0} \right\rangle \cr & \left. a \right){\text{Find 2}}{\bf{u}} \cr & 2{\bf{u}} = 2\left\langle { - 1,2} \right\rangle \cr & 2{\bf{u}} = \left\langle { - 2,4} \right\rangle \cr & \cr & \left. b \right){\text{Find 2}}{\bf{u}} + 3{\bf{v}} \cr & {\text{2}}{\bf{u}} + 3{\bf{v}} = {\text{2}}\left\langle { - 1,2} \right\rangle + 3\left\langle {3,0} \right\rangle \cr & {\text{2}}{\bf{u}} + 3{\bf{v}} = \left\langle { - 2,4} \right\rangle + \left\langle {9,0} \right\rangle \cr & {\text{2}}{\bf{u}} + 3{\bf{v}} = \left\langle { - 2 + 9,4 + 0} \right\rangle \cr & {\text{2}}{\bf{u}} + 3{\bf{v}} = \left\langle {7,4} \right\rangle \cr & \cr & \left. c \right){\text{Find }}{\bf{v}} - 3{\bf{u}} \cr & {\bf{v}} - 3{\bf{u}} = \left\langle {3,0} \right\rangle - 3\left\langle { - 1,2} \right\rangle \cr & {\bf{v}} - 3{\bf{u}} = \left\langle {3,0} \right\rangle + \left\langle {3, - 6} \right\rangle \cr & {\bf{v}} - 3{\bf{u}} = \left\langle {3 + 3,0 - 6} \right\rangle \cr & {\bf{v}} - 3{\bf{u}} = \left\langle {6, - 6} \right\rangle \cr} $$
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