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: 43

Answer

\[\begin{align} & \mathbf{a}.\left\langle -15,23,-26 \right\rangle \\ & \mathbf{b}.\left\langle -31,49,-31 \right\rangle \\ & \mathbf{c}.\sqrt{141} \\ \end{align}\]

Work Step by Step

\[\begin{align} & \mathbf{a}.\text{ 3}\mathbf{u}+\text{2}\mathbf{v} \\ & \text{3}\mathbf{u}+\text{2}\mathbf{v}=3\left\langle -7,11-8 \right\rangle +2\left\langle 3,-5,-1 \right\rangle \\ & \text{3}\mathbf{u}+\text{2}\mathbf{v}=\left\langle -21,33,-24 \right\rangle +\left\langle 6,-10,-2 \right\rangle \\ & \text{3}\mathbf{u}+\text{2}\mathbf{v}=\left\langle -15,23,-26 \right\rangle \\ & \\ & \mathbf{b}.\text{ 4}\mathbf{u}-\mathbf{v} \\ & \text{4}\mathbf{u}-\mathbf{v}=4\left\langle -7,11-8 \right\rangle -\left\langle 3,-5,-1 \right\rangle \\ & \text{4}\mathbf{u}-\mathbf{v}=\left\langle -28,44-32 \right\rangle -\left\langle 3,-5,-1 \right\rangle \\ & \text{4}\mathbf{u}-\mathbf{v}=\left\langle -28-3,44+5,-32+1 \right\rangle \\ & \text{4}\mathbf{u}-\mathbf{v}=\left\langle -31,49,-31 \right\rangle \\ & \\ & \mathbf{c}.\text{ }\left| \mathbf{u}+3\mathbf{v} \right| \\ & \left| \mathbf{u}+3\mathbf{v} \right|=\left| \left\langle -7,11-8 \right\rangle +3\left\langle 3,-5,-1 \right\rangle \right| \\ & \left| \mathbf{u}+3\mathbf{v} \right|=\left| \left\langle -7,11-8 \right\rangle +\left\langle 9,-15,-3 \right\rangle \right| \\ & \left| \mathbf{u}+3\mathbf{v} \right|=\left| \left\langle 2,-4,-11 \right\rangle \right| \\ & \left| \mathbf{u}+3\mathbf{v} \right|=\sqrt{{{\left( 2 \right)}^{2}}+{{\left( -4 \right)}^{2}}+{{\left( -11 \right)}^{2}}} \\ & \left| \mathbf{u}+3\mathbf{v} \right|=\sqrt{4+16+121} \\ & \left| \mathbf{u}+3\mathbf{v} \right|=\sqrt{141} \\ & \\ & \mathbf{a}.\left\langle -15,23,-26 \right\rangle \\ & \mathbf{b}.\left\langle -31,49,-31 \right\rangle \\ & \mathbf{c}.\sqrt{141} \\ \end{align}\]
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