Chapter 7, Problem 69RE

Solution

a.

To calculate: The coordinate of the rotated point in the $x\text{'}y\text{'}$ -coordinate system.

The point is $\left(\frac{3}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)$ .

Given Information:

The point for rotation is $\left(1,2\right)$ .

Calculation:

Consider the given point,

  $\left(1,2\right)$

The rotation points are defined as,

  $x\text{'}=x\cos \alpha +y\sin \alpha$

And,

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

Where $\alpha$ is representing the angle.

Substitute the given point $\left(1,2\right)$ and $\alpha =45°$ .

  $\begin{array}{c}\left(x\text{'},y\text{'}\right)=\left(1\cos 45°+2\sin 45°,-1\cos 45°+2\sin 45°\right)\\ =\left(\frac{1}{\sqrt{2}}+\frac{2}{\sqrt{2}},-\frac{1}{\sqrt{2}}+\frac{2}{\sqrt{2}}\right)\\ =\left(\frac{3}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)\end{array}$

Therefore, the obtained point is $\left(\frac{3}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)$ .

b.

To calculate: The coordinate of the rotated point in the $xy$ -coordinate system.

The point is $\left(0.71,2.12\right)$ .

Given Information:

The point for rotation is $\left(1,2\right)$ .

Calculation:

Consider the given point,

  $\left(1,2\right)$

The rotation points are defined as,

  $x\text{'}=x\cos \alpha +y\sin \alpha$

And,

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

Where $\alpha$ is representing the angle.

Substitute the given point $\left(1,2\right)$ and $\alpha =45°$ .

  $1=x\cos 45°+y\sin 45°$

And,

  $2=-x\sin 45°+y\cos 45°$

Rewrite the obtained system in the form of $AX=B$ .

  $A=\left[\begin{array}{cc}\cos 45°& \sin 45°\\ -\sin 45°& \cos 45°\end{array}\right]$ , $X=\left[\begin{array}{c}x\\ y\end{array}\right]$ , $B=\left[\begin{array}{c}1\\ 2\end{array}\right]$

Solve the equation $AX=B$ .

  $\begin{array}{c}X=A^{-1}B\\ ={\left[\begin{array}{cc}\cos 45°& \sin 45°\\ -\sin 45°& \cos 45°\end{array}\right]}^{-1}\left[\begin{array}{c}1\\ 2\end{array}\right]\\ =\frac{1}{{\cos }^{2}45°+{\sin }^{2}45°}\left[\begin{array}{cc}\cos 45°& -\sin 45°\\ \sin 45°& \cos 45°\end{array}\right]\left[\begin{array}{c}1\\ 2\end{array}\right]\\ =\left[\begin{array}{cc}\frac{1}{\sqrt{2}}& -\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{array}\right]\left[\begin{array}{c}1\\ 2\end{array}\right]\\ =\left[\begin{array}{c}0.71\\ 2.12\end{array}\right]\end{array}$

Therefore, the obtained point is $\left(0.71,2.12\right)$ .