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Currently I am learning about linear algebra and planes, but most tutorials on "how to find the plane equation given a dot and normal vector" assume that I already understand the equation $$Ax + By + Cz = D$$ I've never seen this formula before, and I don't understand how this equation represents a plane. From the Internet I understand that $A, B, C$ are components of the normal vector. I also believe this equation is somewhat similar to a line in $R^2$, which is $$Ax + By = C$$

For this second equation I can solve for $y$ and plot the formula. Plotting a 3D plane doesn't strike me as easy though. This stops me from building up intuition for this equation: Changing the values of $m$ or $b$ in $y = ax + b$ changes the slope and the intercept of the line. Conversely, I don't understand the behaviour of each constant in the plane equation.

What effect will changing the value of $A$ have on the plane equation? And why does the plane equation use the normal vector's components as constants?

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First of all, if don't have some dedicated software (such as Mathematica or Maple) you can plot your plane here (and you right - fiddling with the constants is a great way to build up intuition!).

All right, so the first thing to note is that the equation for the plane is linear in its variables, i.e., no single variable is taken to some power, or put into a trigonometric function, etc., it's just there with its constant... This is what makes the plane flat!

A constant (e.g. $A$) associated with a variable (e.g. $x$) indicates how much the value of the dependent variable (for instance $z$) grows when you change the variable ($x$), just like a normal 2D function $y=Ax$, i.e., $A$ is the slope for that direction.

And as is the case with $B$ in the case of $y=Ax+B$, the constant $D$ simply raises (if it is positive) the entire plane.

I hope that helps. If you have any doubts, let me know in the comments.

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  • $\begingroup$ Thanks for the answer. You lost me at "This is what makes the plane flat!". How do constants imply a flat plane? $\endgroup$ Feb 15, 2016 at 21:20
  • $\begingroup$ No problem! It isn't the constants that achieve this effect, but that the variables ($x,y,z$) are "standing alone", which means that they are not "mixed" with the other variables (for instance something like $x/y+2z=0$ wouldn't be a plane), and that they are linear, i.e. $Ax$ instead of, say, $Ax^3$ or $A \log(x)$. $\endgroup$ Feb 15, 2016 at 21:41
  • $\begingroup$ Ah I see. But how come these constants in the plane equation are the components of the normal vector of the plane? I still don't understand the relationship between the variables and the constants A, B and C. I do understand D of course, since that is the same with a 2d line. $\endgroup$ Feb 15, 2016 at 21:57
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    $\begingroup$ Just understanding the equation itself is one thing. I've explained that in my answer. The gist is this: If we arrange the plane so $z(x,y)=Ax+By+D$, $A$ tells you how much $z$ grows when we change $x$, just like a normal straight line. Same for $B$ and $y$. $\endgroup$ Feb 15, 2016 at 22:06
  • $\begingroup$ @user1534664 (cont.) Another thing is the relationship to the normal vector; the normal vector is the gradient, and the gradient is the vector of partial derivatives... and when you take the derivative of one of the terms, say $Ax$, it just becomes $A$. That's why the normal vector is just the constants: The variables have been derived away. $\endgroup$ Feb 15, 2016 at 22:12
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A very easy intuition of a plane equation is with 2 things:

1) A point on the plane, say $P_0=(x_0 , y_0 , z_0) $

2) The normal vector, perpendicular to each line on plane surface, say $N=<A,B,C>$

Then for any point on the plane $P=(x,y,z)$, $|PP_0|$ vector is on the surface of the plane, so it is perpendicular to the normal vector. That is the dot product is 0.

$|PP_0| = (x-x_0, y-y_0, z-z_0)$

$|PP_0|.N = A*(x-x_0) + B*(y-y_0) + C*(z-z_0) = 0$

Here you have the plane equation $Ax + By + Cz + D = 0$

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  • $\begingroup$ I understand the derivation as this is explained in most tutorials, but I don't understand the equation itsself. How would I be able to create a flat plane perpendicular to the zy plane for example? $\endgroup$ Feb 15, 2016 at 22:02
  • $\begingroup$ @user1534664 A plane perpendicular to zy-plane could be xy-plane. Also xz-plane is perpendicular to it. You should narrow down your criteria. $\endgroup$
    – crbah
    Feb 15, 2016 at 22:06
  • $\begingroup$ true, sorry. I meant parallel to the zy plane* $\endgroup$ Feb 15, 2016 at 22:08
  • $\begingroup$ @user1534664 Then its easy. A plane paralel to zy-plane has norm N=<1,0,0> (unit vector along x direction). And a point on that plane can be (0,0,0). (here i assume your plane is zy-plane) Now you can apply the formula. $\endgroup$
    – crbah
    Feb 15, 2016 at 22:10
  • $\begingroup$ Thanks, I think messing with examples might be the best way to gain intuition. I'm going to keep practicing, and see how that works out. If I have any more questions I might come back. $\endgroup$ Feb 15, 2016 at 22:37
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$(D/A,0,0)$ is the intersection of the plane with $x$-axis, so if you change $A$ you change this intersection and simulanously hould intersectionpoints with $y$- and $z$-axis, just the same for a line in $\mathbb{R}^2$.

You can imagine a plane as the set of vectors, with the same projektion on some normal vector $(A, B, C)$. Projection is given by scalar product, so you get $$Ax+By+Cz=\text{const.}$$

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