Question

chapter ii.8
score: 1/6 answered: 1/6
question 1
determine \\(\angle r_pr_q\\) where \\(r_p\\) and \\(r_q\\) are rays with tails at \\((0,0)\\) that respectively pass through \\(p\\) and \\(q\\), and \\(p = (\frac{1}{4},\frac{\sqrt{15}}{4})\\) and \\(q = (\frac{16}{19},\frac{\sqrt{105}}{19})\\).

Explanation:

Step1: Recall dot - product formula

The dot - product of two vectors $\vec{u}=(x_1,y_1)$ and $\vec{v}=(x_2,y_2)$ is $\vec{u}\cdot\vec{v}=x_1x_2 + y_1y_2$, and $\vec{u}\cdot\vec{v}=\vert\vec{u}\vert\vert\vec{v}\vert\cos\theta$, where $\theta$ is the angle between the two vectors. First, consider the vectors $\vec{u}$ and $\vec{v}$ corresponding to the rays $R_p$ and $R_q$ with endpoints $p = (\frac{1}{4},\frac{\sqrt{15}}{4})$ and $q=(\frac{16}{19},\frac{\sqrt{105}}{19})$ starting from the origin $(0,0)$.
For $\vec{u}=(\frac{1}{4},\frac{\sqrt{15}}{4})$, $\vert\vec{u}\vert=\sqrt{(\frac{1}{4})^2+(\frac{\sqrt{15}}{4})^2}=\sqrt{\frac{1 + 15}{16}} = 1$.
For $\vec{v}=(\frac{16}{19},\frac{\sqrt{105}}{19})$, $\vert\vec{v}\vert=\sqrt{(\frac{16}{19})^2+(\frac{\sqrt{105}}{19})^2}=\sqrt{\frac{256+105}{361}}=\sqrt{\frac{361}{361}} = 1$.
The dot - product $\vec{u}\cdot\vec{v}=\frac{1}{4}\times\frac{16}{19}+\frac{\sqrt{15}}{4}\times\frac{\sqrt{105}}{19}=\frac{4}{19}+\frac{\sqrt{15\times105}}{76}=\frac{4}{19}+\frac{\sqrt{1575}}{76}=\frac{4}{19}+\frac{15\sqrt{7}}{76}$.

Step2: Calculate the angle

Since $\vec{u}\cdot\vec{v}=\vert\vec{u}\vert\vert\vec{v}\vert\cos\theta$ and $\vert\vec{u}\vert = 1$, $\vert\vec{v}\vert = 1$, we have $\cos\theta=\vec{u}\cdot\vec{v}$.
$\vec{u}\cdot\vec{v}=\frac{1}{4}\times\frac{16}{19}+\frac{\sqrt{15}}{4}\times\frac{\sqrt{105}}{19}=\frac{16}{76}+\frac{\sqrt{1575}}{76}=\frac{16 + 15\sqrt{7}}{76}$.
$\theta=\arccos(\frac{16 + 15\sqrt{7}}{76})\approx\frac{\pi}{6}$ radians.

Answer:

$\frac{\pi}{6}$