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: 38

Answer

$$43$$

Work Step by Step

$$\vec{AB}= \lt 1,-4 \gt -\lt -6,0 \gt =\lt 7, -4 \gt \\ \vec{AC}=\lt 3,1 \gt -\lt -6,0 \gt =\lt 9,1 \gt$$ Now, $$\vec{AB}\times\vec{AC}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\7&-4&0\\9&1&0\end{vmatrix}= 43\hat{k}$$ and $$|\vec{AB}\times\vec{AC}|=\sqrt {0+0+(43)^2}=43$$
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