Chapter 9.3, Problem 65E

To determine

To find: The coordinate of the point in the rotated coordinate system.

Answer to Problem 65E

The coordinate of the point in the rotated system is $\left(2\sqrt{3},\text{ 4}\right)$ .

Explanation of Solution

Given information:

The given angle is $\theta ={30}^{\circ }$ and the coordinate is $\left(1,3\right)$ .

Calculation:

The coordinate of the points in the rotated system is given by

  $x=x\text{'}\cos \theta -y\text{'}\sin \theta$ and $y=x\text{'}\sin \theta +y\text{'}\cos \theta$

where $x\text{'}$ and $y\text{'}$ are the coordinate of the point in the rotated coordinate system.

Here, $x=1$ , $y=3$ and $\theta ={30}^{\circ }$ .

Therefore,

  $\begin{array}{l}1=x\text{'}\cos {30}^{\circ }-y\text{'}\sin {30}^{\circ }\\ 1=x\text{'}\times \frac{\sqrt{3}}{2}-y\text{'}\times \frac{1}{2}\\ 1=\frac{\sqrt{3}x\text{'}-y\text{'}}{2}\\ 2=\sqrt{3}x\text{'}-y\text{'}\mathrm{......}\text{ (1)}\end{array}$ and

  $\begin{array}{l}3=x\text{'}\sin {30}^{\circ }+y\text{'}\cos {30}^{\circ }\\ 3=x\text{'}\times \frac{1}{2}+y\text{'}\times \frac{\sqrt{3}}{2}\\ 3=\frac{x\text{'}+\sqrt{3}y\text{'}}{2}\\ 6=x\text{'}+\sqrt{3}y\text{'}\mathrm{......}\text{ (2)}\end{array}$

Multiply equation (1) by $\sqrt{3}$ and add it to equation (2):

  $\begin{array}{l}\text{(1)}\times \sqrt{3}2\sqrt{3}=\sqrt{3}x\text{'}-\sqrt{3}y\text{'}\\ (2)\underset{\_}{6=x\text{'}+\sqrt{3}y\text{'}}\\ 2\sqrt{3}+6=\sqrt{3}x\text{'}+x\text{'}\end{array}$

Now,

  $\begin{array}{l}\sqrt{3}x\text{'}+x\text{'}=2\sqrt{3}+6\\ x\text{'}\left(\sqrt{3}+1\right)=2\sqrt{3}+6\\ x\text{'}=\frac{2\sqrt{3}+6}{\sqrt{3}+1}\\ x\text{'}=\frac{\left(2\sqrt{3}+6\right)\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\\ x\text{'}=\frac{6-2\sqrt{3}+6\sqrt{3}-6}{3-1}\\ x\text{'}=\frac{4\sqrt{3}}{2}\\ x\text{'}=2\sqrt{3}\end{array}$

Substitute $x\text{'}=2\sqrt{3}$ in equation (1) and solve for $y\text{'}$ :

  $\begin{array}{l}2=\sqrt{3}\left(2\sqrt{3}\right)-y\text{'}\\ y\text{'}=6-2\\ y\text{'}=4\end{array}$

Hence,

The coordinate of the point in the rotated system is $\left(2\sqrt{3},\text{ 4}\right)$ .