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: 45



Work Step by Step

We have $\vec{AB}= -\hat{i}+2\hat{j}$ and $\vec{AC}= -\hat{i}-\hat{j}$. Now, $\vec{AB}\times\vec{AC}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\-1&2&0\\-1&0&-1\end{vmatrix}$$=-2\hat{i}+\hat{j}+2\hat{k}$. Therefore, $|\vec{AB}\times\vec{AC}|=\sqrt {4+1+4}= 3$. The area of the given triangle= $\frac{1}{2}|\vec{AB}\times\vec{AC}|=\frac{3}{2}$
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