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 784: 94


Vector $\mathbf{v}$ is given as $\mathbf{v}=\left( {{x}_{2}}-{{x}_{1}} \right)\mathbf{i}+\left( {{y}_{2}}-{{y}_{1}} \right)\mathbf{j}$.

Work Step by Step

Consider vector $\mathbf{v}$ with initial point ${{P}_{1}}=\left( {{x}_{1}},{{y}_{1}} \right)$ and terminal point ${{P}_{2}}=\left( {{x}_{2}},{{y}_{2}} \right)$. Then the vector $\mathbf{v}$ is equal to the position vector. $\mathbf{v}=\left( {{x}_{2}}-{{x}_{1}} \right)\mathbf{i}+\left( {{y}_{2}}-{{y}_{1}} \right)\mathbf{j}$ Example: Information: Consider the vector $\mathbf{v}$ with the initial point ${{P}_{1}}=\left( 5,3 \right)$ and terminal point ${{P}_{2}}=\left( 8,5 \right)$. Then, the position vector $\mathbf{v}$ is given as $\begin{align} & \mathbf{v}=\left( {{x}_{2}}-{{x}_{1}} \right)\mathbf{i}+\left( {{y}_{2}}-{{y}_{1}} \right)\mathbf{j} \\ & =\left( 8-5 \right)\mathbf{i}+\left( 5-3 \right)\mathbf{j} \\ & =3\mathbf{i}+2\mathbf{j} \end{align}$ The position vector $\mathbf{v}$ is $3\mathbf{i}+2\mathbf{j}$.
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