# Points from an affine subspace with equal distance from given points

Given vector space $\mathbb{R}^3$ with dot product defined as $x \cdot y = 2x_1y_1 + 3x_2y_2 + x_3y_3$ where $x = (x_1,x_2,x_3),y = (y_1,y_2,y_3)$ and given an affine subspace $W: x - y - z - 2 = 0$ .

I need to find all points from W which have the same distance from $p_1 = (0,1,1),\enspace p_2 = (-1,1,0),\enspace p_3 = (0, 0, 0)$.

Am I correct that this will be an intersection of three paraboloids in $\mathbb{R}^3$ ? (Following the fact that parabola is the set of all points having the same distance from a point and a line in $\mathbb{R}^2$)

My approach would be, that I need to find $x = (x,y,z)$ satisfying following conditions:

1) $x \in W$ thus I get first equation $(1, -1, -1) \cdot (x,y,z) = 2$

Since $W$ is a hyperplane in $\mathbb{R}^3$ and distance of point $p_i$ from a hyperplane W is given by formula $$\rho(p_i,W) = \frac{|p_i \cdot n_W - 2|}{||n_W||}$$ where $n_W$ is normal vector of $W$. Im aware that the norm in the denominator is induced by the given dot product.

I get 3 more conditions: $\rho(p_1,W) = \rho(p_2,W) = \rho(p_3,W)$

Im not sure how to put all these 4 conditions all together, can anybody give me some hint? Or is my approach even correct or is there a faster (better) approach than mine?

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Note that since you are not using the standard distance metric. Also, in $\mathbb{R}^3$, the set of points which are equidistant from 2 given points is the perpendicular bisector plane. I'm not sure why you are considering parabolas since lines are not involved. –  Calvin Lin Jun 3 '13 at 14:57

You want a (possibly several, though unlikely) point $A=(a,b,c)$ such that
1. $a-b-c-2 = 0$.
2. $4(a-0)^2+9(b-1)^2+(c-1)^2 = 4(a+1)^2 + 9(b-1)^2 + (c-0)^2 = 4(a-0)^2 + 9(b-0)^2 + (c-0)^2$
1. $-2c+1 = 8a + 4$,
2. $-18b + 9 -2c+1= 0$