Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 7 - Applications of Trigonometry and Vectors - Section 7.4 Vectors, Operations, and the Dot Product - 7.4 Exercises - Page 329: 28

Answer

(a) $\langle -4,-2 \rangle$ (b) $\langle -13,4 \rangle$ (c) $\langle 3,5 \rangle$

Work Step by Step

(a) To find the value of $2\textbf{u}$, we substitute the vector $\textbf{u}$ in the expression and simplify: $2\textbf{u}$ $=2\cdot\langle -2,-1 \rangle$ $=\langle 2(-2),2(-1) \rangle$ $=\langle -4,-2 \rangle$ (b) To find the value of $2\textbf{u}+3\textbf{v}$, we substitute the vectors $\textbf{u}$ and $\textbf{v}$ in the expression and simplify: $2\textbf{u}+3\textbf{v}$ $=2\cdot\langle -2,-1 \rangle+3\cdot\langle -3,2 \rangle$ $=\langle 2(-2),2(-1) \rangle+\langle 3(-3),3(2) \rangle$ $=\langle -4,-2 \rangle+\langle -9,6 \rangle$ $=\langle -4-9,-2+6 \rangle$ $=\langle -13,4 \rangle$ (c) To find the value of $\textbf{v}-3\textbf{u}$, we substitute the vectors $\textbf{u}$ and $\textbf{v}$ in the expression and simplify: $\textbf{v}-3\textbf{u}$ $=\langle -3,2 \rangle-3\cdot\langle -2,-1 \rangle$ $=\langle -3,2 \rangle-\langle 3(-2),3(-1) \rangle$ $=\langle -3,2 \rangle-\langle -6,-3 \rangle$ $=\langle -3-(-6),2-(-3) \rangle$ $=\langle 3,5 \rangle$
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