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: 47


$\frac{\sqrt {21}}{2}$

Work Step by Step

We have $\vec{AB}= -\hat{i}+2\hat{j}$ and $\vec{AC}=\hat{j}-2\hat{k}$. Now, $\vec{AB}\times\vec{AC}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\-1&2&0\\0&1&-2\end{vmatrix}$$=-4\hat{i}-2\hat{j}-\hat{k}$. Therefore, $|\vec{AB}\times\vec{AC}|=\sqrt {16+4+1}= \sqrt {21}$. The area of the given triangle= $\frac{1}{2}|\vec{AB}\times\vec{AC}|=\frac{\sqrt {21}}{2}$
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