Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
Try this, it is the answer to your question. –  user12205 Dec 4 '11 at 23:52
1  
@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. –  Pacciu Dec 5 '11 at 1:51

3 Answers 3

up vote 4 down vote accepted

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 $\frac{5}{3}$.

share|improve this answer
    
Thanks that helps a lot! –  jmendegan Dec 5 '11 at 1:25

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.

share|improve this answer

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 $P\cdot W = 0$ $$ P=\left| \begin{matrix} (x,y,z) & 1 \end{matrix} \right|\cdot \left| \begin{matrix} (a,b,c) & \epsilon\end{matrix} \right| = 0 $$ $$ ax+by+cz+\epsilon =0$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.