17
$\begingroup$

The following show you the whole question.

Find the distance d bewteen two planes \begin{eqnarray} \\C1:x+y+2z=4 \space \space~~~ \text{and}~~~ \space \space C2:3x+3y+6z=18.\\ \end{eqnarray} Find the other plane $C3\neq C1$ that has the distance d to the plane $C2$.

According to the example my teacher gave me, the answer should be : enter image description here

enter image description here

Am I right? However, I do not know what is normal and why there are P(5) and Q($-\frac{1}{2}$).


Thank you for your attention

$\endgroup$
  • 3
    $\begingroup$ Hint: rewrite equation for $C2$ $\endgroup$ – freak_warrior Nov 6 '13 at 14:36
  • $\begingroup$ There is only a meaningful distance between planes in $\mathbb{R}^3$ if they are parallel, so think about that. $\endgroup$ – Tim Ratigan Nov 6 '13 at 14:52
  • $\begingroup$ @TimRatigan: There's a meaningful distance $d(A,B):=\inf_{a\in A,b\in B}d(a,b)$ for any two arbitrary subsets $A,B\subseteq\mathbb{R}^3$. $\endgroup$ – C-Star-W-Star Nov 17 '15 at 15:26
  • $\begingroup$ @Freeze_S I suppose he means that the distance is zero if they are not parallel. $\endgroup$ – JP McCarthy Nov 17 '15 at 15:36
16
+500
$\begingroup$

For a plane defined by $ax + by + cz = d$ the normal (ie the direction which is perpendicular to the plane) is said to be $(a, b, c)$ (see Wikipedia for details). Note that this is a direction, so we can normalise it $\frac{(1,1,2)}{\sqrt{1 + 1 + 4}} = \frac{(3,3,6)}{\sqrt{9 + 9 + 36}}$, which means these two planes are parallel and we can write the normal as $\frac{1}{\sqrt{6}}(1,1,2)$.

Now let us find two points on the planes. Let $y=0$ and $z = 0$, and find the corresponding $x$ values. For $C_1$ $x = 4$ and for $C_2$ $x = 6$. So we know $C_1$ contains the point $(4,0,0)$ and $C_2$ contains the point $(6,0,0)$.

The distance between these two points is $2$ and the direction is $(1,0,0)$. Now we now that this is not the shortest distance between these two points as $(1,0,0) \neq \frac{1}{\sqrt{6}}(1,1,2)$ so the direction is not perpendicular to these planes. However, this is ok because we can use the dot product between $(1,0,0)$ and $\frac{1}{\sqrt{6}}(1,1,2)$ to work out the proportion of the distance that is perpendicular to the planes.

$(1,0,0) \cdot \frac{1}{\sqrt{6}}(1,1,2) = \frac{1}{\sqrt{6}}$

So the distance between the two planes is $\frac{2}{\sqrt{6}}$.

The last part is to find the plane which is the same distance away from $C_2$ as $C_1$ but in the opposite direction. We know the normal must be the same, $\frac{1}{\sqrt{6}}(1,1,2)$. Using this we can write $C_3: x + y+ 2z = a$ and determine $a$. When $y=0, z=0$ we moved from $(4,0,0) \rightarrow (6,0,0)$, so if we move the same distance again we go $(6,0,0) \rightarrow (8,0,0)$ and $(8,0,0)$ is on $C_3$. Therefore, $a = 8$. So finally the equation of the plane is,

$C_3: x + y + 2z = 8$

And we are done :)

$\endgroup$
  • 1
    $\begingroup$ +1 For using elementary geometric reasoning and a minimum of prerequisites. $\endgroup$ – dafinguzman Nov 18 '15 at 3:26
  • $\begingroup$ ...and that has earned you a +500... :) $\endgroup$ – Jesse P Francis Nov 18 '15 at 12:12
  • $\begingroup$ Is the final answer for the first part $\frac{1}{\sqrt{6}}$? I really think that the distance between the 2 planes is $\sqrt{\frac23}$ using the workings by lhf $\endgroup$ – mauna Jan 28 '16 at 19:03
  • 1
    $\begingroup$ @mauna - sorry typo, it was $2(\frac{1}{\sqrt{6}}) = \sqrt{\frac{2}{3}}$. I'll make that edit now $\endgroup$ – j__ Jan 28 '16 at 21:58
  • $\begingroup$ Why did you find the unit normal vector? Can't see non unit vector not work? $\endgroup$ – Scáthach Dec 1 '18 at 13:14
9
$\begingroup$

Parallel planes are level sets of a linear function. In this case, $x+y+2z=c$.

The signed distance of $x+y+2z=c$ to the origin is the normalized algebraic value $$ \frac{c}{\sqrt{1^2+1^2+2^2}}=\frac{c}{\sqrt{6}} $$

Therefore, the unsigned distance between two planes $x+y+2z=c_1$ and $x+y+2z=c_2$ is $$ \frac{|c_1-c_2|}{\sqrt{6}} $$

In your example, $c_1=4$ and $c_2=6$ and so the distance is $$ \frac{2}{\sqrt{6}} $$

The other plane at the same distance to $C_2$ has $c=6+2=8$ and so is given by $x+y+2z=8$.

$\endgroup$
3
$\begingroup$

There is another way of finding the distance. Write $C_1$ and $C_2$ as functions $z=f_i(x,y)$:

$$C_1\,:\,z=f_1(x,y)=2-\frac12 x-\frac12 y\qquad\text{ and}$$

$$C_2\,:\,z=f_1(x,y)=3-\frac12 x-\frac12 y.$$

To find the intersections we assume that $(x,y,z)\in C_1\cap C_2$ and solve the simultaneous equations.

$$\begin{align} 2-\frac12 x-\frac12 y&=3-\frac12 x-\frac12 y \\ \Rightarrow 2&=3, \end{align}$$

a contradiction and so $C_1\cap C_2$ is empty and so $C_1\parallel C_2$.

Find a point on one plane, say $(0,0,2)$ on $C_2$.

Write a function $d(x,y)=\operatorname{dist}((x,y,f_1(x,y)),(0,0,2))$ using Pythagoras a couple of times:

$$d(x,y)=\sqrt{x^2+y^2+(1-x/2-y/2)^2}.$$

Minimise $d(x,y)$ --- or even easier $(d(x,y))^2$ --- using partial differentiation:

$$\begin{align} \frac{\partial d^2}{\partial x}&=2x+2(1-x/2-y/2)(-1/2) \\&=\frac52 x -1+\frac{y}{2},\text{ and} \\ \frac{\partial d^2}{\partial y}&= \frac52 y -1+\frac{x}{2}. \end{align} $$

Solve both partial derivatives equal to zero to find that the minimum of $d^2$ occurs at $(x,y)=(1/3,1/3)$. Plug these into $d(x,y)$ to find

$$d_\min=\sqrt{\frac23}.$$

For $C_3$ to be parallel to $C_1$ and $C_2$ it must be of the form:

$$f_3(x,y)=\lambda-\frac12 x-\frac12 y,$$

for $\lambda\neq 2,3$.

Note that

$$f_1(x,y)=-x/2-y/2+2,$$ and $$f_2(x,y)=-x/2-y/2+3=f_1(x,y)+1,$$

and so $C_2$ is just $C_1$ shifted upwards by one. Therefore shift upwards again by one and you will find $C_3$:

$$f_3(x,y)=-x/2-y/2+4\Leftrightarrow x+y+2z=8.$$

You wouldn't even have to solve the first part to do this.

$\endgroup$
2
$\begingroup$

Notice the two planes are parallel. . A point has to be found out on any one of the plane where $y=x=0$. So distance between parallel planes is given by $\frac{ax_1+by_1+cz_1+d}{\sqrt(a^2+b^2+c^2)}$. So it'll be $\frac{ax_1+by_1+cz_1+d}{\sqrt6}$

$\endgroup$
  • $\begingroup$ Since $ax_1+by_1+cz_1=-d_1$, could that also be expressed as $\frac{\left\lvert d - d_1\right\rvert}{\sqrt{a^2+b^2+c^2}}$? $\endgroup$ – Solomon Ucko Apr 25 '19 at 22:48
2
$\begingroup$

The term "normal" means perpendicularity. A normal vector of a plane in three-dimensional space points in the direction perpendicular to that plane. (This vector is unique up to non-zero multiplies.)

For determining the distance between the planes $C_1$ and $C_2$ you have to understand what is the projection of a vector onto the direction of a second one, see https://en.wikipedia.org/wiki/Vector_projection.

The the projection of any vector $\vec{QP}$ pointing from one of the planes to the other onto the direction of the common normal vector $\vec{n}=(1,1,2)$ (or $\vec{n}=(3,3,6)$ because length does not matter but only direction does) is a vector perpendicular to both planes and pointing from one to another. So its length has to be the distance.

Finding $C_3$ is easy as well. $C_3$ is the image of $C_1$ under reflection about $C_2$. So $C_3$ is the set of points $(x,y,z)$ which can be written as $2(x'',y'',z'')-(x',y',z')$ with $(x'',y'',z'')$ in $C_2$ and $(x',y',z')$ in $C_1$. Checking what is $3$(first coordinate)$+3$(second coordinate)$+6$(third coordinate) for the points of $C_3$, we get $3(2x''-x')+3(2y''-y')+6(2z''-z')=2(3x''+3y''+6z'')-3(x'+y'+z')=2\cdot18-3\cdot 4=12$. Hence the equation for the plane $C_3$ is $3x+3y+6z=24$ or $x+y+2z=8$.

$\endgroup$
1
$\begingroup$

Take one point from one plane, e.g. your $P$ from above. You know that the shortest distance between 2 planes (they are parallel, so they have the same normal vector) is the distance of $P$ to the point where it touches the other plane in direction of the outer normal. This means:

construct a linear map

$P+r\cdot n$ where n is the normal vector of the 2 planes.

Then search the point where this map touches the other plane. The point you get from this is the point that has shortest distance to $P$.

As they are parallel, the initial choice of $P$ is not important.

EDIT: You can write a plane as $P + r\cdot v_1 + s v_2$ where $P$ is some(!) point on the plane and $v_1,v_2$ are the direction-vectors of the plane. As you are in $\mathbb{R}^3$, it is always possible to find another vector that is orthogonal to both direction-vectors. This vector is the normal-vector of the plane. Given your plane in coordinate form, the normal vector can directly be read: It consists of the coefficients of your variables, so $3x+4y-2z= 15$ has the normal-vector $\begin{bmatrix}3\\4\\-2\end{bmatrix}$

$\endgroup$
1
$\begingroup$

As others observed, the planes must be parallel.

Suppose you were given $0x + 0y + z = a$ and $0x + 0y + z = b$. Then obviously the distance between them is $d = |a-b|$, and $z = b \pm d$ give the two planes a distance $d$ away from the $z=b$ plane.

So all you need to do is choose your axes to line up with the planes. Let $w$ be the coordinate along the direction $\vec{n} = (1,1,2)$, since that's the form both planes take. But then the coordinate $w = (x,y,z) \cdot \vec{n}/|\vec{n}|$:

$$x + y + 2z = 4 \iff w|\vec{n}| = 4 \iff w = 4/\sqrt{6} $$

and similarly

$$3x + 3y + 6z = 18 \iff 3w|\vec{n}| = 18 \iff w = 6/\sqrt{6} $$

Now you can finish off the problem using the observations in the second paragraph.

$\endgroup$
0
$\begingroup$

This is a sort of geometric way to see this, suppose instead of the finding the distance between the planes $C_1$ and $C_2$, we consider the simpler problem of finding the distance between the planes

$z=2$ and $z=4$

It is clear, in this case that the distance between these planes is $2$ and is realized by the points $(0,0,2)$ and $(0,0,4)$ respectively. Notice, that these points are the intersection of the line parametrized by $(0,0,t)$ and the planes themselves. What's special about this line, is that it is the line paramaterized by the unit vector $[0,0,1]$.

Now, if we turn our heads to the current problem $C_{1}$ and $C_{2}$, we can essentially use the same argument by tilting our heads and treating the normal vector the planes $[1,1,2]$ as the $z$-axis above. Doing this, we see that the intersection of the line $(t,t,2t)$ and plane $C_{1}$ is $(\frac{2}{3}, \frac{2}{3}, \frac{4}{3})$ and the intersection with $C_{2}$ is $(1,1,2)$. Computing the distance between these two points we obtain $\frac{\sqrt{2}}{\sqrt{3}}$ or $\frac{2}{\sqrt{6}}$.

The reason why this works is that we are using $[1,1,2]$ and the plane $x+y+2z=0$ through the origin as a basis for $\mathbb{R}^{3}$. It's a change of coordinates of sorts.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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