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how can I find the geometric location on $\mathbb{R}^3$ for which the distance to $2x-y+2z-6=0$ is twice the distance to $x+2y-2z+3=0$ ?

The equation I'm used to using to compute the distance from point to plane is

$$ d = \frac{| \ Ax + By + Cz + d \ |}{|\vec{n}|} $$

if $\vec{n}$ is the normal vector of the plane and $P = (x,y,z)$ the point.

The answer should be two planes, but I'm having a hard time starting the exercise. Any tips are appreciated.

Thank you.

bru1987
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1 Answers1

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This is more a comment, but it's too long, so I'm making it an answer.

To get an idea of what happens, I'm going to look at the simplest possible versions of this.

For 2D, consider the lines $L: x=0$ and $M: y=0$.

If a point (in the first quadrant) $(x, y)$ is twice as far from $L$ as from $M$, then $x = 2y$, so this is a straight line.

For 3D, consider the two planes $L: x=0$ and $M: y = 0$. If $(x, y, z)$ is twice as far from $L$ as from $M$, then $x = 2y$ and this is a plane through the line $x=y=0$.

So, the result seems plausible.

I'll let someone else to the actual algebra.

marty cohen
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