# same distance between two NON parallel planes?

I have the following problem:

Given two planes $$P_1:\:\:3x+2y-6z=1$$$$P_2:\:\:-3y+4z=3$$Find all points $$v=(a,b,c)$$ such that $$\inf\{d(v,p_1)\mid p_1\in P_1\}=\inf\{d(v,p_2)\mid p_2\in P_2\}$$.

The first thing I checked was that they weren't parallel, so my guess is that I have to find the "plane in-between" that would represent all the points at the same distance between the two planes. but I really don't know how to do it.

Sorry for my English and thanks for reading!

answering amd : (Do you know a formula for the distance of a point to a plane? Write that for an arbitrary point in ℝ3 for each plane and equate them) this is what happens: !

applying it...

!

I don't know how to solve further.

• for two intersecting lines in the plane, the equidistant points requested are on the two angle bisectors Commented Nov 21, 2019 at 4:17
• Do you know a formula for the distance of a point to a plane? Write that for an arbitrary point in $\mathbb R^3$ for each plane and equate them.
– amd
Commented Nov 21, 2019 at 10:24
• Here's the approach proposed by @amd. Note that the shortest distance between a point and a plane is on the line normal to the plane. So, given plane $ax+by+cz=d$ and point $(x_0,y_0,z_0)$, we can find the point on the plane closest to the point by solving $$a(x_0+at)+b(y_0+bt)+c(z_0+ct)=d$$for $t$, and then computing the point $(x_0+at,y_0+bt,z_0+ct)$, and then finding the distance. Once we finish computing the distance, it turns out to be $$d((x_0,y_0,z_0),ax+by+cz=d)=\frac{ax_0+by_0+cz_0-d}{\sqrt{a^2+b^2+c^2}}$$ Commented Nov 21, 2019 at 21:38
• Note that @WillJagy's approach is probably much faster, but also requires a bit more insight into the problem which doesn't seem to be there right now. Commented Nov 21, 2019 at 21:40
• @DonThousand You need to square both sides to get the complete solution.
– amd
Commented Nov 22, 2019 at 2:37

Addendum: If $$\ 1+ 3p_1-2p_2+6p_3\$$ and $$\ 3+3p_2-4p_3\$$ are of the same sign, then your final equation becomes $$1+ 3p_1-2p_2+6p_3=\frac{7}{5}\left(3+3p_2-4p_3\right)\ ,$$ the equation of one plane, while if they're of opposite sign, the equation becomes $$1+ 3p_1-2p_2+6p_3=-\frac{7}{5}\left(3+3p_2-4p_3\right)\ ,$$ the equation of a second plane.
Alternatively, you could follow the suggestion amd made in his or her second comment, and square both sides to get $$\left(1+ 3p_1-2p_2+6p_3\right)^2=-\frac{49}{25}\left(3+3p_2-4p_3\right)^2\ ,$$ which is the equation of a degenerate conic consisting of the same two intersecting planes.