Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 13 - Vector Geometry - 13.4 The Cross Product - Exercises - Page 677: 35

Answer

See the verification below.

Work Step by Step

We need to verify that $||\textbf{v}\times\textbf{w}||^{2}=||\textbf{v}||^{2}||\textbf{w}||^{2}-(\textbf{v}\cdot\textbf{w})^{2}$ $\textbf{v}\times\textbf{w}=\begin{vmatrix}\textbf{i}&\textbf{j}&\textbf{k}\\3&-2&2\\4&-1&2\end{vmatrix}$ $=\textbf{i}(-2\times2-2\times-1)-\textbf{j}(3\times2-4\times2)+\textbf{k}(-1\times3-4\times-2)$ $=-2\textbf{i}+2\textbf{j}+5\textbf{k}$ $||\textbf{v}\times\textbf{w}||=\sqrt {(-2)^{2}+2^{2}+5^{2}}=\sqrt {33}$ $||\textbf{v}\times\textbf{w}||^{2}=33$ $||\textbf{v}||=\sqrt {3^{2}+(-2)^{2}+2^{2}}=\sqrt {17}$ $||\textbf{w}||=\sqrt {4^{2}+(-1)^{2}+2^{2}}=\sqrt {21}$ $\textbf{v}\cdot\textbf{w}=(3\times4)+(-2\times-1)+(2\times2)=18$ $(\textbf{v}\cdot\textbf{w})^{2}=18^{2}=324$ $||\textbf{v}||^{2}||\textbf{w}||^{2}-(\textbf{v}\cdot\textbf{w})^{2}=17\times21-324=33=||\textbf{v}\times\textbf{w}||^{2}$ Hence, verified.
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