I am having trouble with this:

Find the distance from the point $(1,1,1)$ to the plane $2x+2y+z=0$.

Any ideas? Thanks.

  • 1
    $\begingroup$ Try this, it is the answer to your question. $\endgroup$
    – user12205
    Dec 4, 2011 at 23:52
  • 1
    $\begingroup$ @Jeroen Vaelen: I'd like to remark that formula $\frac{|ax_0+by_0+cz_0+d|}{\sqrt{a^2+b^2+c^2}}$ for the distance from the point $(x_0,y_0,z_0)$ to the plane $ax+by+cz+d=0$ holds also in a more general setting: in fact one can prove an analogous formula, known as Ascoli's formula (e.g., see: matematicamente.it/forum/…), in linear normed vector spaces of any dimension. $\endgroup$
    – Pacciu
    Dec 5, 2011 at 1:51

3 Answers 3


The family of planes, indexed by $\alpha$ $$ f(x,y,z)=2x+2y+z=\alpha $$ are all parallel, with normal vectors parallel to $\nabla f=(2,2,1)$.

Moving a distance $d$ along the normal means moving $d\frac{(2,2,1)}{|(2,2,1)|}$. This movement changes $\alpha$ by $d\frac{2\cdot2+2\cdot2+1\cdot1}{|(2,2,1)|}=d|(2,2,1)|$. Thus, the distance between two of these planes is $\frac{|\Delta\alpha|}{|(2,2,1)|}=\frac{|\Delta\alpha|}{3}$.

Since $\alpha=0$ for $2x+2y+z=0$ and $\alpha=2x+2y+z=5$ for the plane that contains $(1,1,1)$, we get the distance from $2x+2y+z=0$ to $(1,1,1)$ to be $\frac{5}{3}$.


The shortest distance will be achieved along a line that is perpendicular to the plane.

The normal vector to the plane can be read off the equation: since the plane is $2x+2y+z=0$, the normal vector of the plane is $(2,2,1)$.

That means that the shortest path from $(1,1,1)$ to the plane will be along a line parallel to $(2,2,1)$. That is, you are looking a value of $t$ such that $$(1,1,1) + t(2,2,1)$$ lies in the plane. That will be the point in the plane closest to $(1,1,1)$. And once you know the point of the plane closest to $(1,1,1)$, you can compute the distance by simply using the formula for distance between two points.


I always use 3D homogeneous coordinates for points and planes with the following constructs:

  1. Point $P=\left| \begin{matrix} \vec{p} & \delta\end{matrix} \right|=\left| \begin{matrix} (p_x,p_y,p_z) & \delta \end{matrix} \right| = \left| \begin{matrix} (1,1,1) & 1\end{matrix} \right|$
  2. Plane $W=\left| \begin{matrix} \vec{w} & \epsilon \end{matrix} \right| = \left| \begin{matrix} (a,b,c) & \epsilon\end{matrix} \right| = \left| \begin{matrix} (2,2,1) & 0 \end{matrix} \right|$
  3. Point Plane Distance $h=\dfrac{\vec{p}\cdot\vec{w}+\delta\,\epsilon}{\delta\,|\vec{w}|} = \dfrac{(1,1,1)\cdot(2,2,1)+0}{1\,\sqrt{2^2+2^2+1^2}} = \dfrac{5+0}{1*3}=\frac{5}{3}$

NOTE: that the equation for the plane is $P\cdot W = 0$ $$ P=\left| \begin{matrix} (x,y,z) & 1 \end{matrix} \right|\cdot \left| \begin{matrix} (a,b,c) & \epsilon\end{matrix} \right| = 0 $$ $$ ax+by+cz+\epsilon =0$$


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