Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 12: Vectors and the Geometry of Space - Section 12.4 - The Cross Product - Exercises 12.4 - Page 719: 37



Work Step by Step

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