Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 11 - Vectors and the Geometry of Space - 11.2 Exercises - Page 764: 59

Answer

$${\bf{z}} = \left\langle {\frac{7}{2},3,\frac{5}{2}} \right\rangle $$

Work Step by Step

$$\eqalign{ & {\text{Let }}{\bf{u}} = \left\langle {1,2,3} \right\rangle ,{\text{ }}{\bf{v}} = \left\langle {2,2, - 1} \right\rangle ,{\text{ }}{\bf{w}} = \left\langle {4,0, - 4} \right\rangle \cr & {\text{2}}{\bf{z}} - 3{\bf{u}} = {\bf{w}} \cr & {\text{2}}{\bf{z}} - 3\left\langle {1,2,3} \right\rangle = \left\langle {4,0, - 4} \right\rangle \cr & {\text{Solve and simplify}} \cr & {\text{2}}{\bf{z}} - \left\langle {3,6,9} \right\rangle = \left\langle {4,0, - 4} \right\rangle \cr & {\text{2}}{\bf{z}} = \left\langle {4,0, - 4} \right\rangle + \left\langle {3,6,9} \right\rangle \cr & {\text{2}}{\bf{z}} = \left\langle {7,6,5} \right\rangle \cr & {\bf{z}} = \frac{1}{2}\left\langle {7,6,5} \right\rangle \cr & {\bf{z}} = \left\langle {\frac{7}{2},3,\frac{5}{2}} \right\rangle \cr} $$
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