Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 11 - Vectors and Vector-Valued Functions - 11.2 Vectors in Three Dimensions - 11.2 Exercises - Page 778: 39

Answer

\[\begin{align} & \mathbf{a}.\left\langle 12,-7,2 \right\rangle \\ & \mathbf{b}.\left\langle 16,-13,-1 \right\rangle \\ & \mathbf{c}.5 \\ \end{align}\]

Work Step by Step

\[\begin{align} & \mathbf{u}=\left\langle 4,-3,0 \right\rangle ,\text{ }\mathbf{v}=\left\langle 0,1,1 \right\rangle \\ & \mathbf{a}.\text{ 3}\mathbf{u}+\text{2}\mathbf{v} \\ & \text{3}\mathbf{u}+\text{2}\mathbf{v}=3\left\langle 4,-3,0 \right\rangle +2\left\langle 0,1,1 \right\rangle \\ & \text{3}\mathbf{u}+\text{2}\mathbf{v}=\left\langle 12,-9,0 \right\rangle +\left\langle 0,2,2 \right\rangle \\ & \text{3}\mathbf{u}+\text{2}\mathbf{v}=\left\langle 12+0,-9+2,0+2 \right\rangle \\ & \text{3}\mathbf{u}+\text{2}\mathbf{v}=\left\langle 12,-7,2 \right\rangle \\ & \\ & \mathbf{b}.\text{ 4}\mathbf{u}-\mathbf{v} \\ & \text{4}\mathbf{u}-\mathbf{v}=4\left\langle 4,-3,0 \right\rangle -\left\langle 0,1,1 \right\rangle \\ & \text{4}\mathbf{u}-\mathbf{v}=\left\langle 16,-12,0 \right\rangle -\left\langle 0,1,1 \right\rangle \\ & \text{4}\mathbf{u}-\mathbf{v}=\left\langle 16-0,-12-1,0-1 \right\rangle \\ & \text{4}\mathbf{u}-\mathbf{v}=\left\langle 16,-13,-1 \right\rangle \\ & \\ & \mathbf{c}.\text{ }\left| \mathbf{u}+3\mathbf{v} \right| \\ & \left| \mathbf{u}+3\mathbf{v} \right|=\left| \left\langle 4,-3,0 \right\rangle +3\left\langle 0,1,1 \right\rangle \right| \\ & \left| \mathbf{u}+3\mathbf{v} \right|=\left| \left\langle 4,-3,0 \right\rangle +\left\langle 0,3,3 \right\rangle \right| \\ & \left| \mathbf{u}+3\mathbf{v} \right|=\left| \left\langle 4,0,3 \right\rangle \right| \\ & \left| \mathbf{u}+3\mathbf{v} \right|=\sqrt{{{\left( 4 \right)}^{2}}+{{\left( 0 \right)}^{2}}+{{\left( 3 \right)}^{2}}} \\ & \left| \mathbf{u}+3\mathbf{v} \right|=\sqrt{16+0+9} \\ & \left| \mathbf{u}+3\mathbf{v} \right|=5 \\ \end{align}\]
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