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:
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applying it…
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I don't know how to solve further.
Best Answer
You're almost there already! But your guess that you need to find the "plane in-between" is only half correct. There are actually two planes "in-between". Can you see how you might derive the equations of two planes (which are actually perpendicular to each other) from your last equation?
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.