# finding esquidistant points from a plane

So, i wanna find points on a plane that are equidistants from :

A=(1,1,0) B=(0,1,1)

π : y = x


what i tried: Set x = 0 -> y=0, z=0.

N(pi) = (1,-1,0)
P1 = (0,0,0);

Dist from pi to A = ((1*1-1*1+0*0) / sqrt(1+1+0) ) = 0;


so i try to figure out what is the distance from pi to B.

Dist from pi to B = ((1*0+(-1)*(-1)+0*1)/sqrt(1+1+0)) = 1/sqrt(2).

but i dont know how to discover the points that are equidistants from A and B from pi.

Thanks.

-

So, you can conclude at least that $A\in\pi$. Now if a point $X$ is equidistant from $A$ and the plane $\pi$, then $AX$ has to be the distance segment itself, hence orthogonal to $\pi$, i.e. $X=A+\lambda\,(1,-1,0)$. (Its distance from $A$ is $|\lambda|\sqrt2$.)

Now $\vec{BX}=A+\lambda\,(1,-1,0)-B=(1+\lambda, -\lambda,-1)$. Its length is $\sqrt{(1+\lambda)^2+\lambda^2+1}$, so we need \begin{align} (1+\lambda)^2+\lambda^2+1 &=2\lambda^2 \\ 2+2\lambda &=0 \\ \lambda &=-1\,. \end{align}

-
the distance from A is λ√2, based on 1/√2? i dont understand this part only. – Matheus Silva Nov 17 '13 at 1:03
$\vec{AX}=X-A=\lambda\cdot(1,-1,0)$. And the length of the vector $(1,-1,0)$ is $\sqrt{1+1+0}=\sqrt2$. – Berci Nov 17 '13 at 2:22