Precalculus (6th Edition) Blitzer

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

Chapter 6 - Section 6.7 - The Dot Product - Exercise Set - Page 799: 88



Work Step by Step

Addition and subtraction for the given vectors $\mathbf{v}={{a}_{1}}\mathbf{i}+{{b}_{1}}\mathbf{j}$ and $\mathbf{w}={{a}_{2}}\mathbf{i}+{{b}_{2}}\mathbf{j}$ is given by $\begin{align} & \mathbf{v}+\mathbf{w}=\left( {{a}_{1}}+{{a}_{2}} \right)\mathbf{i}+\left( {{b}_{1}}+{{b}_{2}} \right)\mathbf{j} \\ & \mathbf{v}-\mathbf{w}=\left( {{a}_{1}}-{{a}_{2}} \right)\mathbf{i}+\left( {{b}_{1}}-{{b}_{2}} \right)\mathbf{j} \\ \end{align}$ Here, ${{a}_{1}}=1,{{a}_{2}}=-2,{{b}_{1}}=-5,{{b}_{2}}=7$ So, $\begin{align} & \mathbf{w}-\mathbf{v}=\left( -2\mathbf{i}+7\mathbf{j} \right)-\left( \mathbf{i}-5\mathbf{j} \right) \\ & =\left( -2-1 \right)\mathbf{i}+\left[ 7-\left( -5 \right) \right]\mathbf{j} \\ & =-3\mathbf{i}+12\mathbf{j} \end{align}$ Hence, the vector $\mathbf{w}-\mathbf{v}=-3\mathbf{i}+12\mathbf{j}$.
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