Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 12 - Vectors and the Geometry of Space - 12.4 Exercises - Page 838: 19

Answer

$\lt (\frac{-1}{3\sqrt 3}),(\frac{-1}{3\sqrt 3}),(\frac{5}{3\sqrt 3}) \gt $ and $\lt (\frac{1}{3\sqrt 3}),(\frac{1}{3\sqrt 3}),(\frac{-5}{3\sqrt 3}) \gt $

Work Step by Step

$\begin{vmatrix} i&j&k \\ 3& 2&1\\-1&1&0\end{vmatrix}= \lt -1,-1,5 \gt$ $d=\sqrt {(-1)^2+(-1)^2+(5)^2}$ $=\sqrt {27}$ $=3\sqrt 3$ $ =\lt -1,-1,5 \gt \div 3\sqrt 3$ $=\lt (\frac{-1}{3\sqrt 3}),(\frac{-1}{3\sqrt 3}),(\frac{5}{3\sqrt 3}) \gt $ $\begin{vmatrix} i&j&k \\ -1&1&0\\ 3& 2&1 \end{vmatrix}= \lt 1,1,-5 \gt$ $d=\sqrt {(1)^2+(1)^2+(-5)^2}$ $=\sqrt {27}$ $=3\sqrt 3$ $ =\lt 1,1,-5 \gt \div 3\sqrt 3$ $=\lt (\frac{1}{3\sqrt 3}),(\frac{1}{3\sqrt 3}),(\frac{-5}{3\sqrt 3}) \gt $
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