Precalculus (6th Edition) Blitzer

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

Chapter 6 - Test - Page 800: 15


a) The desired vector is $\text{v}=\text{i +2j}$ b) The value of $\left\| \text{v} \right\|$ is $\sqrt{5}$.

Work Step by Step

(a) Let us represent the given two points as: $\begin{align} & \left( {{x}_{1}},{{y}_{1}} \right)=\left( -2,3 \right)\, \\ & \,\left( {{x}_{2}},{{y}_{2}} \right)=\left( -1,5 \right) \end{align}$. A position vector between two points $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ denoted in vector form is: $\mathbf{v}\text{ = }\left( {{x}_{2}}-{{x}_{1}} \right)\mathbf{i}+\left( {{y}_{2}}-{{y}_{1}} \right)\mathbf{j}$ …… (1) Now substitute the value of $\left( {{x}_{1}},{{y}_{1}} \right)\,\,\text{ and }\,\,\left( {{x}_{2}},{{y}_{2}} \right)$ in equation (1), $\begin{align} & \mathbf{v}\text{ = }\left( -1+2 \right)\mathbf{i}+\left( 5-3 \right)\mathbf{j} \\ & \text{= }\mathbf{i}+\text{2}\mathbf{j} \end{align}$ (b) The vector v can be computed as: $\mathbf{v}=\mathbf{i}+2\mathbf{j}$ Here, $\left\| \mathbf{v} \right\|$ represents the magnitude of the vector, which is denoted by the formula: $\left\| \mathbf{v} \right\|=\sqrt{{{x}^{2}}+{{y}^{2}}}$ Now in $\mathbf{v}=\mathbf{i}+2\mathbf{j}$ ; $\begin{align} & x=1\, \\ & y=2 \end{align}$ Then, $\begin{align} & \left\| \text{v} \right\|=\sqrt{{{1}^{2}}+{{2}^{2}}} \\ & =\sqrt{5} \end{align}$
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