University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 11 - Practice Exercises - Page 639: 51


$\lt 0,-3,3 \gt$

Work Step by Step

Since, we have $n=\lt 2,1,-1 \gt $ and $v=\lt 1,1,1 \gt$ Thus, the cross product is: $n \times v=i(-1+1)-j(2+1)+k(2+1)=-3j+3k$ Hence, we have $\lt 0,-3,3 \gt$
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