Given points $P(0, 3, 0) \;\;\; Q(-3, 4, 2) \;\;\; R(-2, 9, 1) \;$ find the measure of ∠PQR

Question

I'm having trouble understanding what this is asking for. I'm able to to see that the dot product between vectors PQ and RP = $0$ and that means those sides form a right triangle, but I don't understand what the question wants me to find and I can't find examples of it anywhere.

Here's what I have so far (bold = vector): $$\mathbf a= \mathbf {PQ} \;\;\; and \;\;\; \mathbf c= \mathbf {RP}$$ so $$\mathbf{a} = \left\langle -3,1,0 \right\rangle \;\;\; and \;\;\; \mathbf{c}=\left\langle 2,-6,-1 \right\rangle$$ now $$\mathbf a \cdot\! \mathbf c=0$$ and that means it is a right triangle, but I'm still stuck at what the question is asking for and even then, since I can't find examples anywhere, I'm not even sure I'll understand how to solve it once I figure what it is aksing for.

If someone can explain what is asking for that would be very helpful, and if you feel comfortable with it, even give a hint on how to solve it.

Thank you.

Answer 1

$\vec{QP}=P-Q=\langle 3,-1,-2\rangle$ and $\vec{QR}=R-Q=\langle 1,5,-1\rangle$ so $\langle\vec{QP},\vec{QR}\rangle=0$ and so $\angle PQR=\frac{\pi}{2}=90^\circ$.

Answer 2

Note that for the angle in P

$$\color{red}{\mathbf{a} = \left\langle -3,1,2 \right\rangle} \;\;\; and \;\;\; \mathbf{c}=\left\langle 2,-6,-1 \right\rangle$$

and

$$\cos \theta_P = \frac{\mathbf{a}\cdot \mathbf{c}}{|\mathbf{a}||\mathbf{c}|}\neq 0$$

Otherwise for the angle in Q

$$\mathbf{RQ}={\mathbf{b} = \left\langle -1,-5,1 \right\rangle}$$

and

$$\cos \theta_Q = \frac{\mathbf{a}\cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|}= 0$$

Thus the triangle is a right triangle in Q.