Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 12: Vectors and the Geometry of Space - Section 12.4 - The Cross Product - Exercises 12.4 - Page 719: 43



Work Step by Step

$$\vec{AB}= 6\hat{i}-5\hat{j} \\\vec{AC}= 11\hat{i}-5 \hat{j}$$ The area of the triangle is: $A=\dfrac{1}{2}|\vec{AB}\times\vec{AC}|$ $\vec{AB}\times\vec{AC}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\6&-5&0\\11&-5&0\end{vmatrix} \\ =(-30+55)\hat{k} \\= 25\hat{k}$ and $|\vec{AB}\times\vec{AC}|=\sqrt {25^{2}}= 25$ Now, $Area=\dfrac{25}{2}$
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