Find the measure of ∠PRQ, with points $P(0, 3, 0) \;\;\; Q(-3, 4, 2) \;\;\; R(-2, 9, 1) \;$

Question

I'm trying to solve the problem in the title but I'm having loads of trouble doing so. I'm not sure what angle the question is asking for. Even if I did, I'm not completely sure how to go about it.

As you can tell, this has to do with multivariable calculus, and all I have learned up to this point is about the dot and some cross product rules, as well as some of the material to do with angles and such.

If anyone can get me started, then maybe I'll actually be able to take something away from answering this question.

Sorry that I do not have any work to show, but I am very lost on this question. Thanks for any help.

Answer 1

We can calculate the angle by using the properties of dot products. Since $R$ is the vertex of the angle, $\angle PRQ = \Theta$ is between $\overrightarrow{RP}$ and $\overrightarrow{RQ}$.

$$\overrightarrow{RP}=\pmatrix{2\\-6\\-1}$$ $$\overrightarrow{RQ}=\pmatrix{-1\\-5\\1}$$

And by definition

$$\begin{align} \overrightarrow{RP} \cdot \overrightarrow{RQ} &= \left\lvert \overrightarrow{RP} \right\rvert \left\lvert \overrightarrow{RQ} \right\rvert \cos\Theta \\[2ex] \pmatrix{2\\-6\\-1} \cdot \pmatrix{-1\\-5\\1} &= \left\lvert \pmatrix{2\\-6\\-1} \right\rvert \left\lvert \pmatrix{-1\\-5\\1} \right\rvert \cos\Theta \\[2ex] (2)(-1) + (-6)(-5) + (-1)(1) &= \sqrt{2^2+6^2+1^1} \sqrt{1^2+5^2+1^1} \cos\Theta \\[2ex] -2+30-1 &= \sqrt{41} \sqrt{27} \cos\Theta \\[2ex] \frac{27}{\sqrt{41}\sqrt{27}} &= \cos\Theta \\[2ex] 52.042º &= \Theta = \angle PRQ \end{align}$$

Answer 2

This is done using the definition of the dot product $a.b=|a||b|cos(\theta)$.

First we need to calculate the vectors $\overrightarrow{RP}$ and $\overrightarrow{RQ}$.

$\overrightarrow{RP}=p-r=(0,3,0)-(-2,9,1)=(2,-6,-1)$

$\overrightarrow{RQ}=q-r=(-3,4,2)-(-2,9,1)=(-1,-5,1)$

Now for the dot product $\overrightarrow{RP}.\overrightarrow{RQ}=2*-1+-6*-5+-1*1=27$ and the absolute values $|\overrightarrow{RP}|=\sqrt{2^{2}+(-6)^{2}+(-1)^{2}}=\sqrt{41}$ and $|\overrightarrow{RQ}|=\sqrt{(-1)^{2}+5^{2}+1^{2}}=3\sqrt{3}$.

Finally we substitute these values into the definition so $\theta=cos^{-1}(\frac{27}{3\sqrt{3}\sqrt{41}})=52.042 ^{\circ}$.