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 779: 42

Answer

\[\begin{align} & \mathbf{a}.\left\langle -9,2,8 \right\rangle \\ & \mathbf{b}.\left\langle -23,-1,7 \right\rangle \\ & \mathbf{c}.5\sqrt{2} \\ \end{align}\]

Work Step by Step

\[\begin{align} & \mathbf{u}=\left\langle -5,0,2 \right\rangle ,\text{ }\mathbf{v}=\left\langle 3,1,1 \right\rangle \\ & \mathbf{a}.\text{ 3}\mathbf{u}+\text{2}\mathbf{v} \\ & \text{3}\mathbf{u}+\text{2}\mathbf{v}=3\left\langle -5,0,2 \right\rangle +2\left\langle 3,1,1 \right\rangle \\ & \text{3}\mathbf{u}+\text{2}\mathbf{v}=\left\langle -15,0,6 \right\rangle +\left\langle 6,2,2 \right\rangle \\ & \text{3}\mathbf{u}+\text{2}\mathbf{v}=\left\langle -9,2,8 \right\rangle \\ & \\ & \mathbf{b}.\text{ 4}\mathbf{u}-\mathbf{v} \\ & \text{4}\mathbf{u}-\mathbf{v}=4\left\langle -5,0,2 \right\rangle -\left\langle 3,1,1 \right\rangle \\ & \text{4}\mathbf{u}-\mathbf{v}=\left\langle -20,0,8 \right\rangle -\left\langle 3,1,1 \right\rangle \\ & \text{4}\mathbf{u}-\mathbf{v}=\left\langle -20-3,0-1,8-1 \right\rangle \\ & \text{4}\mathbf{u}-\mathbf{v}=\left\langle -23,-1,7 \right\rangle \\ & \\ & \mathbf{c}.\text{ }\left| \mathbf{u}+3\mathbf{v} \right| \\ & \left| \mathbf{u}+3\mathbf{v} \right|=\left| \left\langle -5,0,2 \right\rangle +3\left\langle 3,1,1 \right\rangle \right| \\ & \left| \mathbf{u}+3\mathbf{v} \right|=\left| \left\langle -5,0,2 \right\rangle +\left\langle 9,3,3 \right\rangle \right| \\ & \left| \mathbf{u}+3\mathbf{v} \right|=\left| \left\langle 4,3,5 \right\rangle \right| \\ & \left| \mathbf{u}+3\mathbf{v} \right|=\sqrt{{{\left( 4 \right)}^{2}}+{{\left( 3 \right)}^{2}}+{{\left( 5 \right)}^{2}}} \\ & \left| \mathbf{u}+3\mathbf{v} \right|=\sqrt{16+9+25} \\ & \left| \mathbf{u}+3\mathbf{v} \right|=\sqrt{50} \\ & \left| \mathbf{u}+3\mathbf{v} \right|=5\sqrt{2} \\ \end{align}\]
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