Question

chapter ii.7
score: 1/5 answered: 3/5
question 4
determine the angle required to rotate (8/9, √17/9) around (0,0) to (16/17, √33/17).

Explanation:

Step1: Recall rotation - matrix formula

For a rotation of a point $(x,y)$ around the origin $(0,0)$ by an angle $\theta$, the new - point $(x',y')$ is given by

$$\begin{pmatrix}x'\\y'\end{pmatrix}=\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}$$

. Let $x = \frac{8}{9}$, $y=\frac{\sqrt{17}}{9}$, $x'=\frac{16}{17}$, and $y'=\frac{\sqrt{33}}{17}$.
We know that $x'=x\cos\theta - y\sin\theta$ and $y'=x\sin\theta + y\cos\theta$. Another way is to use the dot - product formula for vectors. Consider the vectors $\vec{v}=(\frac{8}{9},\frac{\sqrt{17}}{9})$ and $\vec{u}=(\frac{16}{17},\frac{\sqrt{33}}{17})$.
The dot - product of two vectors $\vec{v}=(v_1,v_2)$ and $\vec{u}=(u_1,u_2)$ is $\vec{v}\cdot\vec{u}=v_1u_1 + v_2u_2$ and $\vec{v}\cdot\vec{u}=\vert\vec{v}\vert\vert\vec{u}\vert\cos\theta$, where $\vert\vec{v}\vert=\sqrt{v_1^{2}+v_2^{2}}$ and $\vert\vec{u}\vert=\sqrt{u_1^{2}+u_2^{2}}$.
First, calculate the magnitudes:
$\vert\vec{v}\vert=\sqrt{(\frac{8}{9})^{2}+(\frac{\sqrt{17}}{9})^{2}}=\sqrt{\frac{64 + 17}{81}} = 1$
$\vert\vec{u}\vert=\sqrt{(\frac{16}{17})^{2}+(\frac{\sqrt{33}}{17})^{2}}=\sqrt{\frac{256+33}{289}} = 1$

Step2: Calculate the dot - product

$\vec{v}\cdot\vec{u}=\frac{8}{9}\times\frac{16}{17}+\frac{\sqrt{17}}{9}\times\frac{\sqrt{33}}{17}=\frac{128+\sqrt{561}}{153}$
Since $\vec{v}\cdot\vec{u}=\vert\vec{v}\vert\vert\vec{u}\vert\cos\theta$ and $\vert\vec{v}\vert = 1$, $\vert\vec{u}\vert = 1$, we have $\cos\theta=\frac{128+\sqrt{561}}{153}$
$\theta=\arccos(\frac{128 + \sqrt{561}}{153})\approx0.25$ radians

Answer:

$\arccos(\frac{128+\sqrt{561}}{153})$ radians