Chapter 11, Problem 37RE

To determine

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

Answer to Problem 37RE

The direction angle is $\theta =120°$ .

Explanation of Solution

Given information:

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

Calculation:

Calculate the direction angle,

  $\begin{array}{l}\tan \theta =\frac{opposite}{adjacent}\\ \end{array}$

Or,

  $\tan \theta =\frac{y}{x}$

Where, $y$ is the length of $y$ component and $x$ is the length of $x$ component of the vector.

The figure-(1) is showing vector,

  

Figure-(1)

Here $y=\sqrt{3}$ and $x=-1$ so

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

As $x$ is negative and $y$ is positive in second quadrant only that is always shown as angle ( $\left(180°-\theta \right)$

Therefore, the direction angle is $\theta =120°$ .