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: 43



Work Step by Step

We have $\vec{AB}= 6\hat{i}-5\hat{j}$ and $\vec{AC}= 11\hat{i}-5 \hat{j}$. Area of the triangle = $\frac{1}{2}|\vec{AB}\times\vec{AC}|$ Now, $\vec{AB}\times\vec{AC}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\6&-5&0\\11&-5&0\end{vmatrix}$$=(-30+55)\hat{k}= 25\hat{k}$. Therefore, $|\vec{AB}\times\vec{AC}|=\sqrt {25^{2}}= 25$. Required area=$ \frac{1}{2}\times25=\frac{25}{2}$
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