Chapter 9.3, Problem 66E

To determine

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

Answer to Problem 66E

The coordinate of the point in the rotated system is $\left(\sqrt{3}+\frac{1}{2},\frac{\sqrt{3}}{2}-1\right)$ .

Explanation of Solution

Given information:

The given angle is $\theta ={60}^{\circ }$ and the coordinate is $\left(1,\text{ 2}\right)$ .

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, $x=1$ , $y=2$ and $\theta ={60}^{\circ }$ .

Therefore,

  $\begin{array}{l}1=x\text{'}\cos {60}^{\circ }-y\text{'}\sin {60}^{\circ }\\ 1=x\text{'}\times \frac{1}{2}-y\text{'}\times \frac{\sqrt{3}}{2}\\ 1=\frac{x\text{'}-\sqrt{3}y\text{'}}{2}\\ 2=x\text{'}-\sqrt{3}y\text{'}\mathrm{......}\text{ (1)}\end{array}$ and

  $\begin{array}{l}2=x\text{'}\sin {60}^{\circ }+y\text{'}\cos {60}^{\circ }\\ 2=x\text{'}\times \frac{\sqrt{3}}{2}+y\text{'}\times \frac{1}{2}\\ 2=\frac{\sqrt{3}x\text{'}+y\text{'}}{2}\\ 4=\sqrt{3}x\text{'}+y\text{'}\mathrm{......}\text{ (2)}\end{array}$

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

  $\begin{array}{l}\text{(2)}\times \sqrt{3}4\sqrt{3}=3x\text{'}+\sqrt{3}y\text{'}\\ (1)\underset{\_}{2=x\text{'}-\sqrt{3}y\text{'}}\\ \text{ 4}\sqrt{3}+2=3x\text{'}+x\text{'}\end{array}$

Now,

  $\begin{array}{l}3x\text{'}+x\text{'}=\text{4}\sqrt{3}+2\\ 4x\text{'}=\text{4}\sqrt{3}+2\\ x\text{'}=\frac{\text{4}\sqrt{3}+2}{4}\\ x\text{'}=\sqrt{3}+\frac{1}{2}\end{array}$

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

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

Hence,

The coordinate of the point in the rotated system is $\left(\sqrt{3}+\frac{1}{2},\frac{\sqrt{3}}{2}-1\right)$ .