Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 7 - Applications of Trigonometry and Vectors - Section 7.5 - Applications of Vectors - 7.5 Exercises - Page 343: 60

Answer

$-6$

Work Step by Step

Let $u=\left\langle a,b\right\rangle$ and $v=\left\langle c,d\right\rangle$ be two vectors and $k$ be a real number, then: $ku=\left\langle ka,kb\right\rangle$ and $u\cdot v=ac+bd$ Now for $u=\left\langle -2,1\right\rangle$ and $v=\left\langle 3,4\right\rangle$ we have $3v=3\left\langle 3,4\right\rangle=\left\langle 3(3),3(4)\right\rangle=\left\langle 9,12\right\rangle$ and $u\cdot(3 v)=\left\langle -2,1\right\rangle\cdot \left\langle 9,12\right\rangle=(-2)(9)+(1)(12)=-18+12=-6 $ Hence, $u\cdot(3v)=-6$.
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