Given the equation of a plane $P$: $$-x - 2y + 2z = -5$$ and a point $Q$: $$(2, -3, 0)$$
How do you find four points on $P$ which have a distance of 3√10 units away from from $Q$?
I tried using the pythagorean theorem with the point on $P$ that is closest to $Q$ but I haven't really been able to get anywhere with that.
You're pretty close. The point $K$ on the plane which is closest to $Q$ is
$K = Q - d \dfrac{N}{\| N \|} $
where
$ N = \langle -1, -2, 2 \rangle $
and
$ d = \dfrac{ N \cdot Q + 5 }{\| N \|} = \dfrac{ -1(2) -2(-3) +2(0)+ 5 }{ 3 } = 3 $
So
$ K = (2, -3, 0) - \dfrac{3 (-1, -2, 2)}{3} = (3, -1, -2) $
The minimum distance $QK = 3 $, so the radius of the circle on which the required points lie is $ \sqrt{ (3 \sqrt{10})^2 - (3)^2 } = \sqrt{ 90 - 9 } = \sqrt{ 81} = 9 $
Now you need to generate two spanning vectors that are mutually orthogonal and also orthogonal to $N$. One such vector is
$ U = \langle 2, -1, 0 \rangle $
Another vector is given by
$ V = N \times U = \langle -1, -2, 2 \rangle \times \langle 2, -1, 0 \rangle = \langle 2 , 4, 5 \rangle $
We need to normalize $U$ and $V$, so define
$ u = \dfrac{U}{\| U \| } = \dfrac{1}{\sqrt{5}} \langle 2, -1, 0 \rangle $
and
$ v = \dfrac{V}{\| V \| } = \dfrac{1}{3\sqrt{5}} \langle 2, 4, 5 \rangle $
Now the four desired points can be generated as follows
$ P_i = K + 9\cos(t_i) \ u + 9\sin(t_i) \ v , \ i = 1, 2, 3, 4 $
where the $\{ t_i \}$'s are any four distinct real numbers, for example, one could choose
$ t_1 = 0 , \ t_2 = \dfrac{\pi}{2} , \ t_3 = \pi , \ t_4 = \dfrac{3 \pi}{2} $
to generate four points that are equidistant from each other (corners of a square).