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: 36

Answer

$$29$$

Work Step by Step

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