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: 44

Answer

\[\begin{align} & \mathbf{a}.\left\langle -8,-18,4\sqrt{2} \right\rangle \\ & \mathbf{b}.\left\langle -18,-35\sqrt{3},9\sqrt{2} \right\rangle \\ & \mathbf{c}.3 \\ \end{align}\]

Work Step by Step

\[\begin{align} & \mathbf{u}=\left\langle -4,-8\sqrt{3},2\sqrt{2} \right\rangle ,\text{ }\mathbf{v}=\left\langle 2,3\sqrt{3},-\sqrt{2} \right\rangle \\ & \mathbf{a}.\text{ 3}\mathbf{u}+\text{2}\mathbf{v} \\ & \text{3}\mathbf{u}+\text{2}\mathbf{v}=3\left\langle -4,-8\sqrt{3},2\sqrt{2} \right\rangle +2\left\langle 2,3\sqrt{3},-\sqrt{2} \right\rangle \\ & \text{3}\mathbf{u}+\text{2}\mathbf{v}=\left\langle -12,-24\sqrt{3},6\sqrt{2} \right\rangle +\left\langle 4,6\sqrt{3},-2\sqrt{2} \right\rangle \\ & \text{3}\mathbf{u}+\text{2}\mathbf{v}=\left\langle -12+4,-24\sqrt{3}+6\sqrt{3},6\sqrt{2}-2\sqrt{2} \right\rangle \\ & \text{3}\mathbf{u}+\text{2}\mathbf{v}=\left\langle -8,-18,4\sqrt{2} \right\rangle \\ & \\ & \mathbf{b}.\text{ 4}\mathbf{u}-\mathbf{v} \\ & \text{4}\mathbf{u}-\mathbf{v}=4\left\langle -4,-8\sqrt{3},2\sqrt{2} \right\rangle -\left\langle 2,3\sqrt{3},-\sqrt{2} \right\rangle \\ & \text{4}\mathbf{u}-\mathbf{v}=\left\langle -16,-32\sqrt{3},8\sqrt{2} \right\rangle -\left\langle 2,3\sqrt{3},-\sqrt{2} \right\rangle \\ & \text{4}\mathbf{u}-\mathbf{v}=\left\langle -16-2,-32\sqrt{3}-3\sqrt{3},8\sqrt{2}+\sqrt{2} \right\rangle \\ & \text{4}\mathbf{u}-\mathbf{v}=\left\langle -18,-35\sqrt{3},9\sqrt{2} \right\rangle \\ & \\ & \mathbf{c}.\text{ }\left| \mathbf{u}+3\mathbf{v} \right| \\ & \left| \mathbf{u}+3\mathbf{v} \right|=\left| \left\langle -4,-8\sqrt{3},2\sqrt{2} \right\rangle +3\left\langle 2,3\sqrt{3},-\sqrt{2} \right\rangle \right| \\ & \left| \mathbf{u}+3\mathbf{v} \right|=\left| \left\langle -4,-8\sqrt{3},2\sqrt{2} \right\rangle +\left\langle 6,9\sqrt{3},-3\sqrt{2} \right\rangle \right| \\ & \left| \mathbf{u}+3\mathbf{v} \right|=\left| \left\langle 2,\sqrt{3},-\sqrt{2} \right\rangle \right| \\ & \left| \mathbf{u}+3\mathbf{v} \right|=\sqrt{4+3+2} \\ & \left| \mathbf{u}+3\mathbf{v} \right|=\sqrt{9} \\ & \left| \mathbf{u}+3\mathbf{v} \right|=3 \\ \end{align}\]
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.