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As part of an exercise three equations defining planes are given, of which two were very similar:

$$ \left\{ \begin{array}{c} x + 5y - 2z = 5 \\ -x + 5y + 2z = -4 \end{array} \right. $$

I saw that adding the left hand sides eliminated all variables except $y$:

$$ ( x + 5y - 2z ) + ( -x + 5y + 2z ) = 0x + 10y + 0z = 10y$$ $$ 5 - 4 = 1$$ So: $$10y = 1 \implies y = 0.1$$ Which gives: $$ x + 0.5 -2z = 5 \implies x -2z = 4.5$$ $$ -x + 0.5 + 2z = -4 \implies -x + 2z = -4.5 \implies x - 2z = 4.5$$

I concluded the two equations defined the same plane, but the solution to the exercise given by the book implies three different planes. Where did I make a mistake?


The full exercise:

Do these three planes intersect in (1) a point, (2) a line, (3) two lines (if two planes are parallel), (4) in three parallel lines:

$$ \left\{ \begin{array}{c} x + 5y - 2z = 5 \\ 2x - 4z = 1 \\ -x + 5y + 2z = -4 \end{array} \right. $$

The book gives (4) as the solution, but if the first and third equation simplify to the same $x−2z=4.5$, there would be no intersection, since that is a parallel plane to $2x - 4z = 1$.

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    $\begingroup$ 1. What is the 3rd equation? 2. You've concluded that the above 2 planes intersect somewhere in a plane that's parallel to the $xz$-plane where $y=0.1$. This is not the same as saying they're the same plane. $\endgroup$ – Xoque55 Sep 4 '16 at 22:16
  • $\begingroup$ @Xoque55 1. I added the full exercise, including the third equation, 2. Could you expand on that? Are you saying $y$ does not necessarily have to be $0.1$? Obviously I made a mistake, but I'm not sure what went wrong. $\endgroup$ – 11684 Sep 4 '16 at 22:25
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You just found parametric equations of the intersection line of the two planes: $$\begin{bmatrix} x\\y\\z \end{bmatrix}=\begin{bmatrix}\frac92\\\frac1{10}\\0 \end{bmatrix}+t\begin{bmatrix}2\\0\\1 \end{bmatrix}.$$ The two equations cannot represent the same plane, since they have non-collinear normal vectors: $$\begin{bmatrix}1\\5\\-2 \end{bmatrix}\enspace\text{and}\enspace\begin{bmatrix}-1\\5\\2 \end{bmatrix}.$$

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  • $\begingroup$ The first part of your answer is unfortunately not within my current understanding, but the second half I also noticed. Yet adding the two equations seemed to yield a value for y which, when substituted made the equations equal, so I made a mistake. But what was the mistake? $\endgroup$ – 11684 Sep 4 '16 at 22:29
  • $\begingroup$ It's only a misinterpretation:you've found conditions on the coordinates of a point to belong to both planes (their intersection line), which translates eventually into the parametric equation I mention. Did you not see parametric equations? $\endgroup$ – Bernard Sep 4 '16 at 22:35
  • $\begingroup$ I do not know what parametric equations are, my book doesn't cover that. On the other hand I think I'm starting to make sense of the formula you posted: it describes a set of points (a line because of the linear growth) which goes through $(4.5, 0.1, 0)$ and every $t$ $x$, $y$, and $z$ increase by the corresponding component of the rightmost vector (? my book uses different braces for vectors). Thanks for that, I've been wondering how to represent a line in a 3 dimensional coordinate system. $\endgroup$ – 11684 Sep 4 '16 at 22:44
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    $\begingroup$ The righmost vector is a directing vector of the line, and $t$ is name for the parameter. Actually if $A$ is the point with coordinates $(4.5, 0.1, 0)$ and $M$ is any other point of the line, with coordinates $(x,y,z)$ $\overrightarrow{AM}$ is collinear to the directing vector, and $t $ is the coefficient of proportionality (here, it's equal to $z$). $\endgroup$ – Bernard Sep 4 '16 at 22:55
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You can interpret a plane equation $$ a_x x + a_y y + a_z z = b $$ as scalar product $$ (a_x,a_y,a_z) \cdot (x,y,z) = b $$ where $(a_x,a_y,a_z)$ is a normal vector of the plane and $b$ relates to the signed distance $d$ of the plane to the origin via: $$ d = \frac{b}{\lVert (a_x,a_y,a_z) \rVert} = \frac{b}{\sqrt{a_x^2 + a_y^2 + a_z^2}} $$ This gives $$ (1,5,-2) \cdot (x,y,z) = 5 \\ $$ and $$ (-1, 5, 2) \cdot (x,y,z) = -4 \iff \\ (1, -5, -2) \cdot (x,y,z) = 4 \\ $$ which tells you that these are quite different planes.

Here is an image of the scene, it suggests that the last answer might apply.

scene (Large version)

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  • $\begingroup$ Thank you, the clarification about the relation between $b$ and $d$ is very enlightening. Does that also apply to a 2d line? I.e. $(a_x, a_y) \cdot (x, y) = b \implies d = \frac{b}{\sqrt{a_x^2 + a_y^2}}$? $\endgroup$ – 11684 Sep 4 '16 at 22:53
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    $\begingroup$ Yes, it holds in $n$ dimensions. $\endgroup$ – mvw Sep 4 '16 at 23:21
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I'm not sure what you are doing, probably looking for intersection? Two equations describe the same plane iff they are proportional. This condition should be checked. It is not satisfied.

You may also observe, that the point $(0,1,0)$ lies on the first plane, but not on the second one. This also suffices to prove that the equations describe two different planes. Of course two different planes may intersect or not, this is another question.

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You can simply check if both planes are equal by looking at both equations simultaneously. So we have $$ x+5y−2z-5=−x+5y+2z+4 $$ Reducing that equation we get

$$ 2x-4z-9=0 $$

Since not all variables cancel out those two planes are definitely not the same. They do intersect therefore.

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