Chapter 9, Problem 37RE

To determine

To find: The direction angle between $i$ and $v=-i+\sqrt{3}j$ .

Answer to Problem 37RE

The direction angle between $i$ and $v=-i+\sqrt{3}j$ is $120°$ .

Explanation of Solution

Given information:

The vector is $v=-i+\sqrt{3}j$

Calculation:

The direction angle is given as,

  $\begin{array}{c}\tan \theta =\frac{opposite}{adjacent}\\ \tan \theta =\frac{y}{x}\end{array}$ … (1)

Here, $y$ is the length of the $y$ component and $x$ is the length of the $x$ component.

Draw the graph for the vector $v=-i+\sqrt{3}j$ .

  

From the above graph the value of $y=\sqrt{3}$ and $x=-1$ .

Substitute the values in equation (1).

  $\begin{array}{c}\tan \theta =\frac{\sqrt{3}}{-1}\\ \tan \theta =-\tan 60°\\ \tan \theta =-\tan \left(180°-60°\right)\\ \theta =120°\end{array}$

Therefore, the direction angle between $i$ and $v=-i+\sqrt{3}j$ is $120°$ .