Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 6 - Section 6.6 - Vectors - Exercise Set - Page 782: 56



Work Step by Step

Put $v=6\mathbf{i}-\mathbf{j}$ and $w=-3\mathbf{i}$ in the expression ${{\left\| v+w \right\|}^{2}}-{{\left\| v-w \right\|}^{2}}$. Then, we have $\begin{align} & {{\left\| v+w \right\|}^{2}}-{{\left\| v-w \right\|}^{2}}={{\left\| 6\mathbf{i}-\mathbf{j}+\left( -3\mathbf{i} \right) \right\|}^{2}}-\left\| 6\mathbf{i}-\mathbf{j}-\left( -3\mathbf{i} \right) \right\| \\ & ={{\left\| 6\mathbf{i}-\mathbf{j}-3\mathbf{i} \right\|}^{2}}-{{\left\| 6\mathbf{i}-\mathbf{j}+3\mathbf{i} \right\|}^{2}} \\ & ={{\left\| 3\mathbf{i}-\mathbf{j} \right\|}^{2}}-{{\left\| 9\mathbf{i}-\mathbf{j} \right\|}^{2}} \end{align}$ If $u=a\mathbf{i}+b\mathbf{j}$ is a vector, then its norm is $\left\| u \right\|=\sqrt{{{a}^{2}}+{{b}^{2}}}$. Use this definition of norm to get $\begin{align} & {{\left\| v+w \right\|}^{2}}-{{\left\| v-w \right\|}^{2}}={{3}^{2}}+{{\left( -1 \right)}^{2}}-\left( {{9}^{2}}+{{\left( -1 \right)}^{2}} \right) \\ & =9+1-\left( 81+1 \right) \\ & =10-82 \\ & =-72 \end{align}$ Hence, the given expression is a scalar and its value is $-72$.
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