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: 2

Answer

< 1, -8, -10 >

Work Step by Step

a = <4, 3, -2> b = <2, -1, 1> First find the cross product of vector a and vector b: a x b = | i j k | = | 3 -2| | 4 -2| | 4 3| | 4 3 -2| |-1 1 | i - | 2 1 | j + |2 -1| k | 2 -1 1 | = (3 - 2) i - (4 - (-4)) j + (-4 - 6) k = i - 8j - 10 k = < 1, -8, -10> To prove a x b orthogonal to a, find a dot ( a x b): a⋅(a x b) = < 4, 3, -2> ⋅<1, -8, -10> = 4 + -24 + 20 = 0 Since the dot product is 0, the cross product is orthogonal to vector a because cos (theta) = 0 implies a theta of 90 degrees, thus orthogonal. To prove a x b orthogonal to b, find b dot ( a x b): b⋅(a x b) = < 2, -1, 1> ⋅<1, -8, -10> = 2 + 8 + -10 = 0 Since the dot product is 0, the cross product is orthogonal to vector a because cos (theta) = 0 implies a theta of 90 degrees, thus orthogonal.
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