Chapter 14.2, Problem 20WE

To determine

To find:The angle between $u$ and $v$ by using dot product.

Answer to Problem 20WE

The angle between two given vectors is $94.4$ .

Explanation of Solution

Given information:

There are given two vectors $\overrightarrow{u}=5i-3j$ and $\overrightarrow{v}=2i+4j$ .

Consider the given vectors $\overrightarrow{u}=5i-3j$ and $\overrightarrow{v}=2i+4j$ . 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,

  $\begin{array}{c}\overrightarrow{\left|u\right|}=\sqrt{{\left(5\right)}^2+{\left(-3\right)}^2+{\left(0\right)}^2}\\ =\sqrt{34}\\ \overrightarrow{\left|v\right|}=\sqrt{{\left(2\right)}^2+{\left(4\right)}^2+{\left(0\right)}^2}\\ =\sqrt{20}\end{array}$

Also,

  $\begin{array}{c}\overrightarrow{u}.\overrightarrow{v}=\left(5i-3j\right).\left(2i+4j\right)\\ \overrightarrow{u}.\overrightarrow{v}=5\times 2+\left(-3\right)\times \left(4\right)\\ =-2\end{array}$

Substitute the value of $\overrightarrow{\left|u\right|}=\sqrt{34}$ , $\overrightarrow{v}=\sqrt{20}$ and $\overrightarrow{u}.\overrightarrow{v}=-2$ in the given formula,

  $\begin{array}{c}\cos \theta =\frac{-2}{\sqrt{34}.\sqrt{20}}\\ =\frac{-2}{\sqrt{170}}\\ \theta =94.4\end{array}$

Therefore, the angle between two given vectors is $94.4$ .