Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 9 - Section 9.5 - The Cross Product - 9.5 Exercises - Page 665: 22

Answer

$\sqrt {469}$

Work Step by Step

The cross product is defined as: $u \times v=\begin{vmatrix}i&j&k\\m_1&m_2&m_3\\n_1&n_2&n_3\end{vmatrix}=\lt m_2n_3-m_3n_2, m_3n_1-m_1n_3, m_1n_2-m_2b_1 \gt$ The area of parallelogram $A$ is defined as the length of the cross product $u \times v$ . Now, $u \times v=\lt (-3)(0)-(2)(-6),(2)(5)-(0)(0), (0)(-6) -(-3)(5) \gt=\lt 12,10,15 \gt$ Then, we have $A=|u \times v|=\sqrt {(12)^2+(10)^2+(15)^2}$ or, $A=\sqrt {469}$
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