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: 36



Work Step by Step

We have $\vec{AB}= \lt 7,3 \gt -\lt 0,0 \gt =\lt 7,3 \gt$ and $\vec{AC}=\lt 2,5 \gt -\lt 0,0 \gt =\lt 2,5 \gt$ $\vec{AB}\times\vec{AC}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\7&3&0\\2&5&0\end{vmatrix}=0-0+(35-6)\hat{k}= 29\hat{k}$. Therefore, $|\vec{AB}\times\vec{AC}|=\sqrt {(29)^2}=29$
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