Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 9 - Polar Coordinates; Vectors - 9.7 The Cross Product - 9.7 Assess Your Understanding - Page 631: 6



Work Step by Step

The magnitude of the cross product: $||u\times v||=||u||\cdot||v||\sin{C}.$ Hence the statement is true. By definition the area of a triangle: $K=\frac{uv\cdot \sin{(C)}}{2}$, where $C$ is the included angle of $u$ and $v$. We can compute the area of the parallelogram by adding the areas of the two triangles that make the parallelogram. Which here (according to the picture) is: $\frac{uv\cdot \sin{(C)}}{2}+\frac{uv\cdot \sin{(C)}}{2}=uv\cdot \sin{(C)}=||u||\cdot||v||\cdot \sin{(C)}.$ Therefore the statement is true.
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