Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 12 - Review - Exercises - Page 842: 6

Answer

$ \lt \frac{7}{3 \sqrt 6},\frac{2}{3 \sqrt 6},\frac{-1}{3 \sqrt 6}\gt$, and $ \lt \frac{-7}{3 \sqrt 6},\frac{-2}{3 \sqrt 6},\frac{1}{3 \sqrt 6}\gt$

Work Step by Step

As we know, the cross product of two vectors is orthogonal. We will have to calculate the cross product of the two vectors. Let $a=j+2k$ $\implies$ $a= \lt 0,1,1 \gt$ $b=i-2j+3k$ $\implies$ $b= \lt 1,-2,3 \gt$ $a \times b= \lt 0,1,1 \gt \times \lt 1,-2,3 \gt = \lt 7,2,-1 \gt$ $|a \times b| =\sqrt {49+4+1}= 3 \sqrt 6$ The unit vector is given by: $ \lt \frac{7}{3 \sqrt 6},\frac{2}{3 \sqrt 6},\frac{-1}{3 \sqrt 6}\gt$ The second vectors can be found by reversing the direction of the first. Thus, $ \lt \frac{-7}{3 \sqrt 6},\frac{-2}{3 \sqrt 6},\frac{1}{3 \sqrt 6}\gt$ Hence, $ \lt \frac{7}{3 \sqrt 6},\frac{2}{3 \sqrt 6},\frac{-1}{3 \sqrt 6}\gt$, and $ \lt \frac{-7}{3 \sqrt 6},\frac{-2}{3 \sqrt 6},\frac{1}{3 \sqrt 6}\gt$
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