Question

chapter ii.8
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determine ∠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 $\vec{u}=(u_1,u_2)$ and $\vec{v}=(v_1,v_2)$ is $\vec{u}\cdot\vec{v}=u_1v_1 + u_2v_2$, 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. Let $\vec{u}$ be the vector corresponding to ray $R_p$ with endpoints $(0,0)$ and $p = (\frac{1}{4},\frac{\sqrt{15}}{4})$, and $\vec{v}$ be the vector corresponding to ray $R_q$ with endpoints $(0,0)$ and $q=(\frac{16}{19},\frac{\sqrt{105}}{19})$.

Step2: Calculate the vectors

$\vec{u}=(\frac{1}{4},\frac{\sqrt{15}}{4})$ and $\vec{v}=(\frac{16}{19},\frac{\sqrt{105}}{19})$.

Step3: Calculate the dot - product

$\vec{u}\cdot\vec{v}=\frac{1}{4}\times\frac{16}{19}+\frac{\sqrt{15}}{4}\times\frac{\sqrt{105}}{19}=\frac{4}{19}+\frac{\sqrt{15\times105}}{76}=\frac{4}{19}+\frac{\sqrt{1575}}{76}=\frac{4}{19}+\frac{15\sqrt{7}}{76}=\frac{16 + 15\sqrt{7}}{76}$.

Step4: Calculate the magnitudes

$\vert\vec{u}\vert=\sqrt{(\frac{1}{4})^2+(\frac{\sqrt{15}}{4})^2}=\sqrt{\frac{1 + 15}{16}} = 1$.
$\vert\vec{v}\vert=\sqrt{(\frac{16}{19})^2+(\frac{\sqrt{105}}{19})^2}=\sqrt{\frac{256+105}{361}}=\sqrt{\frac{361}{361}} = 1$.

Step5: Find the cosine of the angle

Since $\vec{u}\cdot\vec{v}=\vert\vec{u}\vert\vert\vec{v}\vert\cos\theta$ and $\vert\vec{u}\vert = 1$, $\vert\vec{v}\vert = 1$, then $\cos\theta=\frac{16 + 15\sqrt{7}}{76}$.
$\theta=\arccos(\frac{16 + 15\sqrt{7}}{76})\approx0.25$ radians.

Answer:

$\arccos(\frac{16 + 15\sqrt{7}}{76})$ radians