Chapter 14.2, Problem 22WE

To determine

To find:The angle between given vectors by using dot product.

Answer to Problem 22WE

The angle between two given vectors is ${90}^{\circ }$ .

Explanation of Solution

Given information:

There is given a vector $\overrightarrow{v}=i+j$ .

Consider the given vectors $\overrightarrow{u}=2i+3j$ and $\overrightarrow{v}=6i-4j$ Then the angle between these two vectors can be calculated as,

  $\cos \theta =\frac{\overrightarrow{u}.\overrightarrow{v}}{\left|\overrightarrow{u}\right|.\left|\overrightarrow{v}\right|}$ .

Now the magnitude of vector can be calculated as,

  $\begin{array}{c}\overrightarrow{\left|u\right|}=\sqrt{{\left(2\right)}^2+{\left(3\right)}^2+{\left(0\right)}^2}\\ \overrightarrow{\left|u\right|}=\sqrt{13}\\ \overrightarrow{\left|v\right|}=\sqrt{{\left(6\right)}^2+{\left(-4\right)}^2+{\left(0\right)}^2}\\ \overrightarrow{\left|v\right|}=\sqrt{52}\end{array}$

Also,

  $\begin{array}{c}\overrightarrow{u}.\overrightarrow{v}=\left(2i+3j\right).\left(6i-4j\right)\\ \overrightarrow{u}.\overrightarrow{v}=2\times 6+3\times \left(-4\right)\\ =0\end{array}$

Substitute the value of $\overrightarrow{u}=\sqrt{13}$ , $\overrightarrow{v}=\sqrt{52}$ and $\overrightarrow{u}.\overrightarrow{v}=0$ in the given formula $\cos \theta =\frac{\overrightarrow{u}.\overrightarrow{v}}{\left|\overrightarrow{u}\right|.\left|\overrightarrow{v}\right|}$ .

  $\begin{array}{c}\cos \theta =\frac{0}{\sqrt{13}.\sqrt{52}}\\ =0\\ \theta =90°\end{array}$

Therefore, the angle between two given vectors is ${90}^{\circ }$ .