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 794: 69


Two vectors are said to be orthogonal vectors if and only if their dot product is zero or they make an angle of ${{90}^{\circ }}$.

Work Step by Step

Consider $\mathbf{v}=2\mathbf{i}+\mathbf{j}$, $\mathbf{w}=2\mathbf{k}$, and $\mathbf{v}\cdot \mathbf{w}=0$. Find the magnitude of vector v as follows: $\begin{align} & \left\| \mathbf{v} \right\|=\sqrt{{{\left( 2 \right)}^{2}}+{{\left( 1 \right)}^{2}}} \\ & =\sqrt{4+1} \\ & =\sqrt{5} \end{align}$ Therefore, $\left\| \mathbf{v} \right\|=\sqrt{5}$. Similarly, find magnitude of vector w as follows: $\begin{align} & \left\| \mathbf{w} \right\|=\sqrt{{{\left( 1 \right)}^{2}}} \\ & =\sqrt{1} \end{align}$ Therefore, $\left\| \mathbf{w} \right\|=1$. Now, find the cosine of the angle between their directions as follows: $\begin{align} & \cos \theta =\left( \frac{\mathbf{v}\cdot \mathbf{w}}{\left\| \mathbf{v} \right\|\text{ }\left\| \mathbf{w} \right\|} \right) \\ & =\left( \frac{0}{\sqrt{5}\left( 1 \right)} \right) \\ & =\left( \frac{0}{\sqrt{5}} \right) \\ & =0 \end{align}$ So, $\begin{align} & \theta ={{\cos }^{-1}}\left( 0 \right) \\ & ={{90}^{\circ }} \end{align}$ Hence, the angle between two vectors is ${{90}^{\circ }}$.Therefore, it can be concluded that the vectors are perpendicular to each other or the vectors are orthogonal.
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