Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 12 - Vectors and the Geometry of Space - Review - Exercises - Page 883: 11


(a) $\lt 4,-3,4 \gt$ (b) $\frac{\sqrt {41}}{2}$

Work Step by Step

(a) $AB= \lt 1,0,-1 \gt$ and $AC= \lt 0,4,3 \gt$ The cross product of the two vectors in the plane will be orthogonal to the plane. $\lt 1,0,-1 \gt \times \lt 0,4,3 \gt= \begin{vmatrix} i&j&k \\1&0&-1& \\0&4&3 \end{vmatrix}$ $=\lt 4,-3,4 \gt$ (b) $Area \triangle ABC=\frac{1}{2}|A|$ $a=AB= \lt 1,0,-1 \gt$ and $b=AC= \lt 0,4,3 \gt$ Here, $A= a \times b= \lt 1,0,-1 \gt \times \lt 0,4,3 \gt$ $= \begin{vmatrix} i&j&k \\1&0&-1& \\0&4&3 \end{vmatrix}$ $A=\lt 4,-3,4 \gt$ $|A|=\sqrt {4^2+(-3)^2+4^2}=\sqrt {41}$ Now, $Area \triangle ABC=\frac{1}{2}\sqrt {41}$ $=\frac{\sqrt {41}}{2}$
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