Chapter 9, Problem 70RE

To determine

To calculate:Thevector $v$ in the $xy$ plane with the magnitude $5$ and the angle between $v$ and $i$ is ${150}^{\circ }$ .

Answer to Problem 70RE

The value of the vectorsis:

  $\begin{array}{l}{v}_1=\frac{-5\sqrt{3}}{2}i+\frac{5}{2}j\\ v_2=\frac{5\sqrt{3}}{2}i-\frac{5}{2}j\end{array}$ .

Explanation of Solution

Given information:

The vectors:

  $v=-2i+j$ and $w=4i-3j$

Formula used:

With the help of unit vector find the vectors.

Such as first find the value of $x$ with the help of angle given in the question and with the help of value of $x$ find the value of $y$ .

Calculation:

Consider the vectors:

  $v=-2i+j$ and $w=4i-3j$

  $\Vert v\Vert =5$ ,

Direction angle, $\alpha ={150}^{\circ }$ .

Now $v=-2i+j$ and $w=4i-3j$ .

Now have to find the vector $v$ ,

  $v=xi+yj$ ,

And the vector,

  $\sqrt{x^2+y^2}=5$

Now the angle:

  $$

  $\begin{array}{c}\tan \alpha =\tan {150}^{\circ }\\ =\frac{y}{x}\end{array}$

Now from the data that is given we have:

  $\begin{array}{c}{x}^2+y^2=25\\ \frac{y}{x}=\frac{5}{2}\end{array}$

Rewrite the expression:

  $\begin{array}{c}{x}^2+y^2=25\\ y=\frac{5}{2}\\ x^2+{\left(\frac{5}{2}\right)}^2=25\\ x^2+\frac{25}{4}{x}^2=25\end{array}$

Now,

  $\begin{array}{c}{x}^2+\frac{25}{4}=25\\ x^2=25-\frac{25}{4}\\ x^2=\frac{100-25}{4}\\ x^2=\frac{75}{4}\end{array}$

Where the value of $x$ is,

  $\begin{array}{l}{x}^2=\frac{75}{4}\\ x=\sqrt{\frac{75}{4}}\\ x=\frac{5\sqrt{3}}{2}\end{array}$

Then the values are:

  $x_1=\frac{-5\sqrt{3}}{2},y_1=\frac{5}{2}$

From this,

  $\begin{array}{c}\tan \alpha =\tan {150}^{\circ }\\ =\frac{y}{x}\\ =\frac{1}{\sqrt{3}}\end{array}$

Here,

  $\frac{y}{x}=\frac{-1}{\sqrt{3}}$

Same as the values of $x_2$ are:

  $x_2=-\frac{5\sqrt{3}}{2},y_2=-\frac{5}{2}$

Then the vectors are:

  $\begin{array}{l}{v}_1=-\frac{5\sqrt{3}}{2}i+\frac{5}{2}j\\ v_2=\frac{5\sqrt{3}}{2}i-\frac{5}{2}j\end{array}$

Thus, the vectors are:

  $\begin{array}{l}{v}_1=-\frac{5\sqrt{3}}{2}i+\frac{5}{2}j\\ v_2=\frac{5\sqrt{3}}{2}i-\frac{5}{2}j\end{array}$ .