University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 11 - Section 11.4 - The Cross Product - Exercises - Page 623: 37

Answer

$$13$$

Work Step by Step

$$\vec{AB}= \lt 2,0 \gt -\lt -1,2 \gt =\lt 3, -2 \gt \\ \vec{AC}=\lt 7,1 \gt -\lt -1,2 \gt =\lt 8,-1 \gt$$ Now, $$\vec{AB}\times\vec{AC}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\3&-2&0\\8&-1&0\end{vmatrix}=(-3+16)\hat{k}= 13\hat{k}$$ and $$|\vec{AB}\times\vec{AC}|=\sqrt {(13)^2}=\sqrt {169}=13$$
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