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The problem

So recently in school, we should do a task somewhat like this (roughly translated):

Assign a system of linear equations to each drawing

Then, there were some systems of three linear equations (SLEs) where each equation was describing a plane in their coordinate form and some sketches of three planes in some relation (e.g. parallel or intersecting at 90°-angles.

My question

For some reason, I immediately knew that these planes:

enter image description here

belonged to this SLE: $$ x_1 -3x_2 +2x_3 = -2 $$ $$ x_1 +3x_2 -2x_3 = 5 $$ $$-6x_2 + 4x_3 = 3$$

And it turned out to be true. In school, we proved this by determining the planes' intersecting lines and showing that they are parallel, but not identical.
However, I believe that it must be possible to show the planes are arranged like this without a lot of calculation. Since I immediately saw/"felt" that the planes described in the SLE must be arranged in the way they are in the picture (like a triangle). I could also determine the same "shape" on a similar question, so I do not believe that it was just coincidence.

What needs to be shown?

So we must show that the three planes described by the SLE cut each other in a way that I do not really know how to describe. They do not intersect with each other perpendicular (at least they don' have to to be arranged in a triangle), but there is no point in which all three planes intersect. If you were to put a line in the center of the triangle, it would be parallel to all planes.

The three planes do not share one intersecting line as it would be in this case:

enter image description here

(which was another drawing from the task, but is not relevant to this question except for that it has to be excluded)

My thoughts

If you were to look at the planes exactly from the direction in which the parallel line from the previous section leads, you would see something like this:

enter image description here

The red arrows represent the normal of each plane (they should be perpendicular). You can see that the normals somehow are part of one (new) plane. This is already given by the manner how the planes intersect with each other (as I described before). If you now were to align your coordinate system in such a way that the plane in which the normals lie is the $x_1 x_2$-plane, each normals would have an $x_3$ value of $0$. If you were now to further align the coordinate axes so that the $x_1$-axis is identical to one of the normals (let's just choose the bottom one), the values of the normals would be somehow like this:

$n_1=\begin{pmatrix} a \\ 0 \\ 0 \end{pmatrix}$ for the bottom normal

$n_2=\begin{pmatrix} a \\ a \\ 0 \end{pmatrix}$ for the upper right normal

and $n_3=\begin{pmatrix} a \\ -a \\ 0 \end{pmatrix}$ for the upper left normal

Of course, the planes do not have to be arranged in a way that the vectors line up so nicely that they are in one of the planes of our coordinate system.

However, in the SLE, I noticed the following:

-The three normals (we can simpla read the coefficients since the equations are in coordinate form) are $n_1=\begin{pmatrix} 1 \\ -3 \\ 2 \end{pmatrix}$, $n_2=\begin{pmatrix} 1 \\ 3 \\ -2 \end{pmatrix}$ and $n_3=\begin{pmatrix} 0 \\ -6 \\ 4 \end{pmatrix}$.

As we can see, $n_1$ and $n_2$ have the same values for $x_1$ and that $x_2(n_1)=-x_2(n_2)$; $x_3(n_1)=-x_3(n_2)$

Also, $n_3$ is somewhat similar in that its $x_2$ and $x_3$ values are the same as the $x_2$ and $x_3$ values of $n_1$, but multiplied by the factor $2$.

I also noticed that $n_3$ has no $x_1$ value (or, more accurately, the value is $0$), while for $n_1$ and $n_2$, the value for $x_1$ is identical ($n_1=1$).

Conclusion

I feel like I am very close to a solution, I just don't know what to do with my thoughts/approaches regarding the normals of the planes.
Any help would be greatly appreciated.

How can I show that the three planes are arranged in this triangular-like shape by using their normals, i.e. without having to calculate the planes' intersection lines? (Probably we will need more than normals, but I believe that they are the starting point).


Update: I posted a new question that is related to this problem, but is (at least in my opinion) not the same question.

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  • $\begingroup$ Is there a typo in the original planes? Should the second one be $x_1 + 3x_2 - 2x_3 = 5$? $\endgroup$ – paulinho Jul 14 at 20:36
  • $\begingroup$ @paulinho Hi! No, the equation is as I wrote it in the question. I might add that to the question if it seemed like a typo. $\endgroup$ – Jonas Jul 14 at 20:38
  • $\begingroup$ I'm not sure if those planes make that picture then. It seems like they intersect in one point. If the second equation was what I wrote above, then indeed the picture is like that $\endgroup$ – paulinho Jul 14 at 20:40
  • $\begingroup$ @paulinho sorry, you were right. I apparently either didn't read your comment or my question properly (I'll better go to sleep now). I've edited my question; it should be correct now. $\endgroup$ – Jonas Jul 14 at 20:58
  • $\begingroup$ Three non-parallel planes will not intersect at a point if the triple scalar product of the normals is zero. $\endgroup$ – Andrew Chin Jul 14 at 21:00
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If you write your systems of equations as a matrix as follows: $$A \vec{x} = \begin{bmatrix} 1 & -3 & 2 \\ 1 & 3 & -2 \\ 0 & -6 & 4 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} -2 \\ 5 \\ 3\end{bmatrix} = \vec{b}$$ then here is a (perhaps) quicker way to determine if the picture looks like the triangle. Note: I don't know how comfortable you are with basic linear algebra concepts, but you only need them to understand the proof of why this is correct. You can apply the method without any understanding of them.

$1$. If all three normal vectors of the planes are multiples of the same vector, then you can immediately conclude you have three parallel planes (and not the triangle).

$2$. If exactly two normal normal vectors are multiples of the same vector, then you can immediately conclude you don't have the triangle. Instead, you have one plane that is cut by two parallel planes.

$3$. If none of the normal vectors are multiples of each other, then it's possible you have the triangle. As you noted, the normal vectors must be in the same plane, i.e. linearly dependent, so it must follow that $\det(A) = 0$. If this isn't the case, then you can immediately conclude that the planes intersect in one point.

$4$. If there is a solution, then $\vec{b}$ should be a linear combination of two linearly independent columns of $A$. (This is because $A \vec{x}$ is just a linear combination of $A$'s columns. If there is a solution to $A \vec{x} = \vec{b}$ and $A$ has two linearly independent columns, then $\vec{b}$ should be able to be written as a linear combination of just those two columns.) Thus, if we replace a linearly dependent column (i.e. one that can be expressed as a linear combination of the others) of $A$ with the vector $\vec{b}$ to create the matrix $A'$, for there to be no solution (i.e. the "triangle" configuration) it must be the case that $\det(A') \neq 0$. If $\det(A') = 0$, then you can conclude you have three planes intersecting in one line (the second picture you've posted).

Fortunately, choosing a linearly dependent column is easy. You just need to make sure to a) replace a zero column with $\vec{b}$ if $A$ has a zero column or b) if there are two columns that are (nonzero) multiples of each other, then replace one of them with $\vec{b}$. And if none of a) or b) is the case, then you can choose any column.

Example: I'll work thru the steps above with the example you've written.

Steps $1$ and $2$. I can immediately notice that none of normal vectors of the planes are parallel. So we proceed to step $3$.

Step $3$. We can calculate $$\det(A) = (1)(12 - 12) - (-3)(4 - 0) + 2(-6 - 0) = 0$$ so we proceed to step $4$. Note that if you were able to observe that the third row of $A$ was a linear combination of the first and second row (the third row is simply the first row minus the second row) or that the third column was a multiple of the second column, you could immediately skip to step $4$.

Step $4$. We can notice that none of the columns are zeroes (case a), but in fact the last two columns are multiples of each other. So case b) applies here, and we have to exchange one of the last two columns with $\vec{b}$ for the process to be correct. Let's replace the last column of $A$ with $\vec{b}$ to obtain $A'$: $$A' = \begin{bmatrix} 1 & -3 & -2 \\ 1 & 3 & 5 \\ 0 & -6 & 3 \end{bmatrix}$$ and we can calculate $$\det (A') = (1)(9 + 30) - (-3)(3 - 0) + (-2)(-6 - 0) = 29 + 9 + 12 = 60 \neq 0$$ and hence we can conclude we have the "triangle" configuration.

Conclusion: I think this method is somewhat easier than calculating the three intersection lines. It requires you to calculate two determinants of $3 \times 3$ matrices instead.

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  • $\begingroup$ Which method did you use to calculate the determinant? On Wikipedia (en.wikipedia.org/wiki/Determinant), I found $det(A)=aei+bfg+cdh-ceg-bdi-afh$ which however seems a bit different from yours (I am however, motly new to linear Algebra, so maybe it is just a misunderstanding of mine) $\endgroup$ – Jonas Jul 18 at 20:24
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    $\begingroup$ @Jonas If you look on the same Wikipedia page on the line above the one you wrote, you see $$\det(A) = a \begin{vmatrix} e & f \\ h& i \end{vmatrix} - b \begin{vmatrix} d & f \\ g& i \end{vmatrix} + c \begin{vmatrix} d & e \\ g & h \end{vmatrix}$$That's the definition I used, but whichever one you use, you should get the same answer. $\endgroup$ – paulinho Jul 18 at 20:44
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The three normals $n_1, n_2, n_3$ all lie in a plane $P$ through the origin, because $n_1 - n_2 = n_3.$ The three given planes are orthogonal to $P.$ If their lines of intersection with $P$ were concurrent, the point of intersection of the lines would lie on all three planes. But if a point $x = (x_1, x_2, x_3)$ is common to the first two planes, then $x \cdot (n_1 - n_2) = x \cdot n_1 - x \cdot n_2 = -2 - 5 = -7,$ which contradicts the equation $x \cdot n_3 = 3$ of the third plane. Therefore the lines of intersection of the three given planes with $P$ are not concurrent. No two of them are parallel, because no two of $n_1, n_2, n_3$ are scalar multiples of each other. Therefore the lines of intersection of the given planes with $P$ intersect each other in three distinct points, forming a triangle in $P.$

(It seems to me that that is all that needs to be said, but I've a horrible feeling I'm missing something $\ldots$)

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  • $\begingroup$ Perhaps it's worth adding that if $p, q, r$ are the lines of intersection of the given planes with $P,$ and $s$ is the line through the origin containing all vectors normal to $P,$ then the three planes are $p + s, q + s, r + s.$ $\endgroup$ – Calum Gilhooley Jul 15 at 13:55
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There is a very easy-to-check necessary and sufficient condition :

You will have the first figure (triangle) if and only if there exists a linear combination of the LHS of your system of equations (1),(2),(3) making $0$ without the RHS being so with the same coefficients ; precisely here :

$$\begin{cases} \text{(condition A)} \ \ & \color{red}{[-1]} \times (1) + \color{red}{[1]} \times (2) + \color{red}{[1]} \times (3) &=& 0 \ \ \text{whereas}\\ \text{(condition B)} \ \ & \color{red}{[-1]} \times -2 + \color{red}{[1]} \times 5 + \color{red}{[1]} \times 3 &\neq & 0\end{cases}$$

We would be in the second case (triangle reduced to $0$ = pencil of planes) iff the RHS is $0$ as well.

Remark:

  1. The proof of this fact, as remarked by you, is that condition A is equivalent to a linear dependency of the normals, whereas condition B amounts to the negation of the fact that for example the 3rd plane is a member of the pencil of planes defined by the first and second plane.

  2. There is a more "linear algebra way" to express remark 1). Let me borrow for that the notations of the excellent answer by @paulinho, working this time with an augmented matrix : $$\exists ? \ \vec{y} \ \text{such that} \ \ \ \underbrace{\begin{bmatrix} y_1 \ \ y_2 \ \ y_3 \end{bmatrix}}_{\vec{y}}\underbrace{[A \ | \ \vec{b}]}_B=\begin{bmatrix} y_1 \ \ y_2 \ \ y_3 \end{bmatrix}\left[\begin{array}{rrr|r} 1 & -3 & 2 & -2 \\ 1 & 3 & -2 & 5 \\ 0 & -6 & 4 & 3 \end{array}\right]=0 $$

Either rank$(B)=3$, no such $\vec{y}$ exists and we are in the first case of the necessary and sufficient condition; otherwise, if rank$(B)<3$ : we are in the second case.

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  • $\begingroup$ Probably a very stupid question - What is an LHS and RHS? $\endgroup$ – Jonas Jul 20 at 20:54
  • $\begingroup$ Left hand side and right hand side of an equation $\endgroup$ – Owen Jul 20 at 21:00
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If you know some linear algebra, this problem becomes easier to describe and answer. The concepts and terminology from linear algebra capture the relevant ideas well.

As long as the null space of the system of three planes has dimension 1, the planes will form a triangle (as in your 1st picture) or intersect in a line (as in your 2nd picture). In simple terms, when I say the null space has dimension 1, I mean that if you changed the right hand side of the system to all 0’s and look at the solutions to this system (called a homogeneous system), you will get a line (through the origin).

The idea is as follows: Each plane $ax + by + cz = d$ is just a translation of the plane $ax + by + cz = 0$. For example, $2x + y - 3z = 4$ is a translation of $2x + y - 3z = 0$. So if the null space has dimension 1, this means the three planes corresponding to 0's on their right sides intersect in a line. So if you translate them back to the original planes of the original system, this just corresponds to moving each plane parallel to its original position. Thus we will either form a triangle (with the three parallel lines) or remain intersecting in a single line.

So, in conclusion, if you have a $3 \times 3$ system of rank 2 that corresponds to the intersection of three planes, e.g. $\left[\begin{array}{rrr|r} 1 & -2 & 3 & 2 \\ 1 & 3 & 4 & 3 \\ 2 & 1 & 7 & 4\end{array}\right]$, then the rank-nullity theorem tells us the null space has dimension 1. (Here by "rank 2" I mean that the non-augmented matrix has rank 2). As we reasoned geometrically above, the only possibilities for the solution set to this system is either it forms a triangle, i.e. no solution, or they intersect in a line, i.e. a line of solutions. To check that they form a triangle instead of intersecting at a line, you can reduce the augmented matrix. If you get an equation like $0 = 1$ in one of the rows then there is no solution, i.e. no point of intersection of the three planes. This is the desired triangle that you asked about. On the other hand if you do not get a row like that, then the system has a solution, so the intersection must be a line.

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  • $\begingroup$ I'm not a mathematician so please excuse anything I have missed. It is possible for three planes to meet at a single point rather than at a single line. Have you covered this possibility? $\endgroup$ – chasly - supports Monica Jul 15 at 13:37
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    $\begingroup$ @chasly Yes of course that is possible. But in that case the null space has dimension 0, i.e. just the zero vector. My question focuses on the specific case when the null space has dimension 1, i.e. it is a line. $\endgroup$ – twosigma Jul 15 at 13:39
  • $\begingroup$ @chasly sorry if my answer was a bit cryptic. I did omit some details. I am also not a mathematician and neither am I a teacher so I might not have explained it very well. $\endgroup$ – twosigma Jul 15 at 14:02
  • $\begingroup$ [+1] I appreciate your explanation in terms of rank. $\endgroup$ – Jean Marie Jul 17 at 13:12
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A line in the Euclidean space can be described by a system of two equations describing a plane.

The equations will be in the form: $$ \begin{cases} ax + by + cz + d = 0 \\ a'x + b'y +c'z + d' = 0 \end{cases} $$ Another way to express a line is in parametric form: $$ \begin{cases} x = x_0 + l\cdot t\\ y = y_0 + m\cdot t \\ z = z_0 + n\cdot t \\ \end{cases} $$ Two lines are parallel if they have the same direction vectors $(l,m,n)$ or if they differ by a scalar multiplication.

You can compute the direction vectors with the formula: $$ (l,m,n) = \left(\begin{vmatrix} b & c \\ b' & c' \end{vmatrix}, -\begin{vmatrix} a & c \\ a' & c' \end{vmatrix}, \begin{vmatrix} a & b \\ a' & b' \end{vmatrix}\right) $$

If you pick any combination of two equations from your example and convert to the parametric form, you will see that they all have the same direction vectors, meaning the intersections between them are parallel.

Also, if you arrange the coefficients into two matrices, like this:

Incomplete matrix $$ A = \begin{pmatrix} a & b & c \\ a' & b' & c' \\ a'' & b'' & c'' \\ a''' & b''' & c''' \end{pmatrix} $$ Complete matrix $$ B = \begin{pmatrix} a & b & c & d \\ a' & b' & c' & d'\\ a'' & b'' & c'' & d''\\ a''' & b''' & c''' & d''' \end{pmatrix} $$

you'll have that the two lines are parallel if the rank of $A$ is 2 and the rank of B is 3.

From the equations above it is easy to see from the coefficients that such matrices would not be full rank because of the repeated terms.

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Good job looking at the normals instead of blindly computing intersections! Indeed the point is that the three vectors $(1, -3, 2)$, $(1, 3, -2)$, $(0,-6, 4)$ are linearly dependent but the vectors $(1, -3, 2, -2)$, $(1, 3, -2, 5)$, $(0,-6, 4, 3)$ are not. But somehow I feel that you may be at a point in your mathematics education where that means nothing to you, so let me put this into a short direct argument.

The sum of the second and third equation is $$x_1 -3x_2 +2x_3 = 8.\qquad(X)$$ So every point which satisfies the second and third equation also satisfies $X$. In other words, the plane described by $X$ contains the intersection of the second and third plane. Now the plane described by $X$ is parallel to the first plane (the left hand sides are the same), but it is not the same plane. That is all we wanted to know. The line of intersection of the second and third plane is parallel to, but not contained in, the first plane.

Oh, now I have talked about intersections after all. Well, there is a duality going on here...

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I guess the reason you "immediately knew" that the system

$$ x_1 -3x_2 +2x_3 = -2 \tag1 \label{eq1}$$ $$ x_1 +3x_2 -2x_3 = 5 \tag2 \label{eq2}$$ $$-6x_2 + 4x_3 = 3 \tag3 \label{eq3}$$

behaved like that

3 planes intersecting, intesections are parallel lines

was that you saw (maybe subconciously) that adding \eqref{eq2} and \eqref{eq3} and subtracting \eqref{eq1} leads to

$$ 0 = 10,$$

showing that there cannot exist a point where all planes intersect.

That can happen in several ways, the most obvious being that 2 of the planes are parallel. But parallel planes are easy to identify in algebraic form, if they are given as

$$a_1x_1+a_2x_2+a_3x_3=z_a$$ $$b_1x_1+b_2x_2+b_3x_3=z_b$$

than being parallel means that there exists a number $f$ such that $b_1=fa_1, b_2=fa_2, b_3=fa_3.$ It's easy to see that this isn't true for any pair of planes described by \eqref{eq1},\eqref{eq2},\eqref{eq3}.

However, that means that each of the 3 pairs of planes have a line as intersection, making 3 lines of intersection. But any two of those lines can't intersect themselves, because that would mean their point of intersection would lie on all 3 planes, which is impossible. Since any 2 lines of intersection lie in of the 3 planes, that means they are parallel!

So we've come to the conclusion that the planes described by \eqref{eq1},\eqref{eq2} and \eqref{eq3} form that picture: They each intersect pairwise, but their intersections are parallel.

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