Calculus: Early Transcendentals 8th Edition

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

Chapter 12 - Section 12.4 - The Cross Product - 12.4 Exercises - Page 821: 20

Answer

$\frac{\sqrt 3} { 3}i-\frac{\sqrt 3} {3}j-\frac{\sqrt 3} {3}k$ and $-\frac{\sqrt 3} { 3}i+\frac{\sqrt 3} {3}j+\frac{\sqrt 3} {3}k$

Work Step by Step

$j-k = \lt 0,1,-1 \gt$ and $i+j = \lt 1,1,0 \gt$ Cross product will produce a vector orthogonal to both. $v=\lt 0,1,-1 \gt \times \lt 1,1,0 \gt = \lt 1,-1,-1 \gt$ $|v|=\sqrt {1^2+(-1)^2+(-1)^2}=\sqrt 3$ $\frac {v}{|v|}=\lt \frac{1} {\sqrt 3},-\frac{1} {\sqrt 3},-\frac{1} {\sqrt 3}\gt$ $=\frac{\sqrt 3} { 3}i-\frac{\sqrt 3} {3}j-\frac{\sqrt 3} {3}k$ The other orthogonal vector is: $=-\frac{\sqrt 3} { 3}i+\frac{\sqrt 3} {3}j+\frac{\sqrt 3} {3}k$
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