Following on from a previous question - Plane Equation Where D Not Equal to Zero:

I am trying to develop an intuitive understanding for the equation of a plane, specifically, plane equations where $D \neq 0$:

$Ax + By + Cz = D$

In the image below is a 3-D space. The orange parallelogram represents a plane which does not intersect the origin. Points $P_1$, $P_2$ and $P_3$ all lie on the plane.

enter image description here

These points can be represented with position vectors; where $\vec v_1$, $\vec v_2$ and $\vec v_3$ correspond to the points $P_1$, $P_2$ and $P_3$ respectively:

enter image description here

Based on how I currently understand it, the vectors pictured below: $(\vec v_1 - \vec v_2)$ and $(\vec v_3 - \vec v_2)$, are parallel to the plane, which is to say that if either of them were moved and placed directly onto the plane, they would lie completely flat across it. Taking the cross product of $(\vec v_1 - \vec v_2) \times (\vec v_3 - \vec v_2)$ should yield a vector ($\vec n$) which is orthogonal to both $(\vec v_1 - \vec v_2)$ and $(\vec v_3 - \vec v_2)$, and therefore the plane. I haven't attempted to draw the normal vector $\vec n$ because it seems to me that it would be around about on the y-axis - so it should be pretty easy to visualise:

enter image description here

Based on what I have just gone through, there are two facts that seem to be in contradiction of each other:

  1. The vector $\vec n$ is normal to both $(\vec v_1 - \vec v_2)$ and $(\vec v_3 - \vec v_2)$, $\vec n$ is therefore normal to the plane (as $(\vec v_1 - \vec v_2)$ and $(\vec v_3 - \vec v_2)$, are parallel to the plane).
  2. $Ax + By + Cz \neq 0$

The value of $D$ in the plane equation cannot be equal to $0$, since the plane does not pass through the origin. Conversely, either of the vectors $(\vec v_1 - \vec v_2)$ or $(\vec v_3 - \vec v_2)$ dotted with the normal vector $(Ax + By + Cz)$ should be equal to $0$ since both vectors are orthogonal to the normal vector (or at least it appears this way to me).

  • $\begingroup$ I don't see how these facts are contradictory. The $D$ cancels when you compute the dot products: $$(A,B,C)\cdot(\vec{v_1}-\vec{v_2}) = (A,B,C)\cdot \vec{v_1}-(A,B,C)\cdot\vec{v_2}=D-D=0.$$ $\endgroup$
    – kccu
    Aug 27, 2019 at 19:35
  • $\begingroup$ Nice diagrams, what program did you use to draw them? $\endgroup$ Aug 27, 2019 at 19:46
  • 1
    $\begingroup$ @electronpusher thanks, I just used Google Drawings. $\endgroup$ Aug 28, 2019 at 16:28

2 Answers 2


There’s no contradiction here. What seems to be tripping you up is that being normal to a plane is not the same thing as being orthogonal to the position vector of a point on the plane. To reduce the proliferation of symbols, in the following I’ll identify points with their position vectors, as is often done when working in $\mathbb R^n$.

We can use your property #1 to define a plane in $\mathbb R^3$: given a vector $\vec n=(A,B,C)\ne0$ and a point $P_0$, the plane through $P_0$ with normal $\vec n$ is the set of points $P$ such that $\vec n\cdot(P-P_0)=0$. It’s easy to verify property #1 for this definition. If $P_1$ and $P_2$ are any two points on the plane, then $$\vec n\cdot(P_1-P_2) = \vec n\cdot(P_1-P_0+P_0-P_2) = \vec n\cdot(P_1-P_0)+\vec n\cdot(P_2-P_0)=0.$$ We can rewrite the defining equation of the plane as $$\vec n\cdot P=\vec n\cdot P_0.\tag{*}$$ The point $P_0$ is fixed, so this equation says that the dot product with $\vec n$ of (the position vector of) every point on the plane is constant. Call this constant $-D$ and expand this equation in terms of coordinates to obtain the familiar $Ax+By+Cz+D=0$. Note that it doesn’t necessarily follow that $\vec n\cdot P=0$.

Now suppose that $P=\alpha\vec n$ is a point on the plane. We then have $\vec n\cdot P = \alpha\vec n\cdot\vec n=-D$, from which $\alpha = -D/\lVert\vec n\rVert^2$. So, if $\alpha\gt0$, we can reach the plane from the origin by moving a distance of $\lvert D\rvert/\lVert\vec n\rVert$ in the direction of $\vec n$; if $\alpha\lt0$, we move in the direction opposite to $\vec n$; and if $\alpha=0$, the origin lies on the plane. Since $\vec n$ is orthogonal to the plane, $\lvert D\rvert/\lVert\vec n\rVert$ the also the distance of the plane from the origin.

Recall that the orthogonal projection of a vector $\vec v$ onto $\vec n$ is given by the expression $$\left({\vec n\over\lVert\vec n\rVert} \cdot \vec v\right){\vec n\over\lVert\vec n\rVert} = {\vec n\cdot\vec v\over \vec n\cdot\vec n}\vec n.$$ Multiplying both sides of equation (*) by $\vec n/(\vec n\cdot\vec n)$, we have $${\vec n\cdot P\over\vec n\cdot\vec n}\vec n = {\vec n\cdot P_0\over\vec n\cdot\vec n}\vec n,$$ which gives us another characterization of the plane: it’s the set of points whose position vectors have the same orthogonal projection onto a fixed nonzero vector $\vec n$. This last should make clear why $\vec n\cdot P$ for a point on the plane is not necessarily equal to zero: this fixed projection onto $\vec n$ can be any scalar multiple of $\vec n$.

  • $\begingroup$ Just to confirm; in your second paragraph, do the points $P$ and $P_0$ both lie on the plane as well as $P_1$ and $P_2$? $\endgroup$ Aug 28, 2019 at 17:49
  • $\begingroup$ In your fifth paragraph, you say - "suppose that $P = \alpha \vec n$ is a point on the plane". Not really sure what you mean here. If I understand you correctly, in the equation $P = \alpha \vec n$ , the normal vector is $\vec n$ and $\alpha$ is a scalar value. If this is in fact what you mean, then $P$ is the point at which the normal vector and the plane intersect? Or have I misinterpreted this? $\endgroup$ Aug 28, 2019 at 18:23
  • $\begingroup$ @RyanWalter Correct on the first count. For the second, it’s the intersection of the plane and the line through the origin parallel to $\vec n$. Vectors aren’t directed line segments, though it’s convenient to visualize them as such, and unless $\alpha\ge1$, that little line segment doesn’t intersect the plane, anyway. $\endgroup$
    – amd
    Aug 28, 2019 at 19:22
  • $\begingroup$ If viewed as position vectors, then $P$ is obviously collinear to $\vec n$ right? $\endgroup$ Aug 29, 2019 at 18:37

There's no contradiction.

The confusion comes about because you seem to think that the $(x,y,z)$ in the plane equation represents any vector parallel to the plane. But that's not the case.

Let us complete your derivation of the equation of the plane. I'll think of points $u,v,w$ as in the plane. Then the displacement vectors $\vec{uv}$ and $\vec{uw}$ are parallel to the plane. Thus, a normal to the plane is $$\mathbf n=\vec{uv}×\vec{uw}.$$ This normal is orthogonal to any vector parallel to the plane. Thus, if $p=(x,y,z)$ represents an arbitrary point in the plane, then an arbitrary vector in the plane is given by $\vec{up},$ so that the equation of the plane is given by $$\mathbf n\cdot \vec{up}=0.$$ Expanding this shows that there is a constant term which comes out nonzero unless of course the plane passes through the origin, which is not so in your case. Thus, you can see that $(x,y,z)$ represents an arbitrary point, and not an arbitrary vector, in the plane.

There's another way to interpret the equation $$ax+by+cz=d$$ which will tell you the way to see this intuitively without first having to put it in the form above. One just uses whichever is convenient for the situation. Namely, $(x,y,z)$ is an arbitrary point in the plane, and we can dually think of it as a position vector. Similarly, $(a,b,c)$ is a certain position vector. The condition says that (by imaginatively dividing both sides by $|(a,b,c)|$ and thinking only of absolute values) the equation describes the set of points with position vector $(x,y,z)$ the length of whose shadow on a fixed position vector $(a,b,c)$ is constant. You can now see that the set of such points is a plane with the fixed vector perpendicular to it. The number $d,$ taken absolutely and divided by $\sqrt {a^2+b^2+c^2}$ now gives the distance of the plane from the origin, that is $$\frac{|d|}{\sqrt {a^2+b^2+c^2}}.$$ It is then clear that the plane passes through the origin if and only if $d$ vanishes. But this has nothing to do with the original interpretation -- it's an alternative one.

In any case you can see that the symbol $(x,y,z)$ does not describe a vector lying in the plane, but it's a point in the plane, or equivalently, the position vector of such a point.

  • $\begingroup$ Could you explain what you mean by - "the length of whose shadow on a fixed position vector $(a,b,c)$ is constant". $\endgroup$ Aug 28, 2019 at 18:05
  • $\begingroup$ @RyanWalter By shadow I meant projection. In other words, that way of interpreting the plane equation says that the plane is the set of position vectors $(x,y,z)$ the length of whose projection onto the fixed position vector $(a,b,c)$ is constant. Draw a picture to verify that this is indeed the case. $\endgroup$
    – Allawonder
    Aug 28, 2019 at 19:21
  • $\begingroup$ Which is to say that no matter which $(x, y, z)$ you pick, the value of $D$ will never change in the plane equation (obviously providing that the point $(x, y, z)$ does lie on the plane)? $\endgroup$ Aug 29, 2019 at 18:25
  • $\begingroup$ @RyanWalter You mean $d$? Well, whenever we say that the distance between a plane and a point is a given value, say $d,$ we mean the length perpendicular segment between the point and the plane. So, what's true is that the length of any all components of position vectors $(x,y,z)$ perpendicular to the plane is constant. $\endgroup$
    – Allawonder
    Aug 29, 2019 at 19:19
  • 1
    $\begingroup$ @RyanWalter Exactly. The constant is $D.$ But in this case $D$ is not the distance of the plane from the origin. It's $|D|$ divided by the length of the fixed vector. $\endgroup$
    – Allawonder
    Aug 30, 2019 at 6:51

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