Chapter 9, Problem 71RE

To determine

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

Answer to Problem 71RE

The value of the vectorsis ${120}^{\circ }$ .

Explanation of Solution

Given information:

The vectors:

  $v=-i+\sqrt{3}j$

Formula used:

Formula to find the angle between two vectors:

  $\cos \theta =\frac{u\cdot v}{\Vert u\Vert \Vert v\Vert }$

Either this is solved by the above formula or the formula given below:

  $\tan \alpha =\frac{y}{x}$

Calculation:

Consider the vectors:

  $v=-i+\sqrt{3}j$

Here, $x=-1,y=\sqrt{3}$ .

To find the direction angle alpha of the given vector $v$ that is $v=-i+\sqrt{3}j$ .

Observe that the expression is given as:

  $v=xi+yj$

Apply the formula for computing the direction angles:

  $\tan \alpha =\frac{y}{x}$

Here the values of $x$ and $y$ are:

  $x=-1,y=\sqrt{3}$

Then to compute the $\alpha$ :

  $\begin{array}{c}\tan \alpha =\frac{y}{x}\\ =\frac{\sqrt{3}}{-1}\\ =-\sqrt{3}\end{array}$

Now the value of $\alpha$ is:

  $\begin{array}{l}\alpha =-\sqrt{3}\\ \alpha =\frac{2\pi }{3}\\ \alpha ={120}^{\circ }\end{array}$

Thus, the direction angle between $i$ and $v$ is: ${120}^{\circ }$ .