Chapter 6, Problem 59CR

To determine

To calculate:

The magnitude and the direction of the vector $v$ .

Answer to Problem 59CR

The magnitude is $\Vert v\Vert =\sqrt{41}$ and the direction angle of the vector $v$ is $\theta =38.65°$ .

Explanation of Solution

Given information:

Vector: $v=5i+4j$

Calculation:

As know that if $z=a+bi,$ then the magnitude of the vector $z$ is $r=\sqrt{a^2+b^2}$ and the direction angle of the vector $z$ is $\tan \theta =\frac{-b}{a}$

Given that $v=5i+4j$

Then find the magnitude and the direction angle of the vector $v$ .

The direction angle of vector $v$ .

  $\begin{array}{l}\tan \theta =\frac{4}{5}\\ =0.8\end{array}$

  $v=5i+4j$ lies in the first quadrant, so $\theta$ also lies in first quadrant

  $\begin{array}{l}\theta ={\tan }^{-1}\left(\tan 38.65°\right)\\ \theta =38.65°\end{array}$

And the magnitude of the vector $v$

  $\begin{array}{l}v=5i+4j\\ \Vert v\Vert =\sqrt{{\left(5\right)}^2+{\left(4\right)}^2}\\ =\sqrt{25+16}\\ =\sqrt{41}\\ \Vert v\Vert =\sqrt{41}\end{array}$

Therefore, the magnitude is $\Vert v\Vert =\sqrt{41}$ and the direction angle of the vector $v\theta =38.65°$