Understanding what a plane is in $\mathbb R^3$

Question

I understand how spheres circles and so on work, My interpretation comes from the sum of their co-ordinates equals the $radius^2$.

I understand how this works but with planes Im really confused. The sum of their $x,y,z$ coordinates equals some value, and that value always makes a square...is that perhaps where I'm going wrong, are planes not always square.

Either way can someone help me get about of intuition on how this formula actually draws out the shape with respect to what the sum of the $x,y,z$ values equal$?$

Answer 1

There are a couple of ways to think about a plane, I don't think a square is a fruitful one, in the same way that a segment isn't a good way to think about a line.

In 2-D, we know that given any $(x_{1},y_{1})$ and $(x_{2},y_{2})$ we can define a line that goes through these two points. We know that this line is unique and fully described just by virtue of going through these two points.

Next, let's talk about a line in 3-D: of course, still, all we need are two points, we can call them $(x_{1},y_1,z_1)$ and $(x_2,y_2,z_2)$. We can still define a line here! But instead of a typical "slope" you need something clearer in 3-dimensions: a vector. So we know that the line that goes through these two points can be totally represented as follows:

$(x_1,y_1,z_1)+t[x_2-x_1,y_2-y_1,z_2-z_1]$. I'm using some scalar $t$, in order to fully describe the line.

Well, 2 points are to a line, as two lines are to a plane. However, these lines are defined parametrically with vectors.

However (I'm assuming that there isn't a lot of linear algebra background): if you wanted to define a line, you couldn't do that with two points that were in the same place. In the same way, you have to make sure that the two vectors you use to define your plane are distinct, or in math-jargon: linearly independent. So you can't have $(x,y,z)+[t1,2,3]$ and $(x,y,z)+s[2,4,6]$ to define a plane, because these generate the same line. So instead:

you look at the cross product of these two vectors: (I'll call the two vectors $\vec n$ and $\vec m$.) Then a linear combination of these vectors fully describes the plane, so they have to lay on the same "flat plane."

So $(x,y,z)=(x_1,y_1,z_1)+t \vec n+s\vec m$

Basically, it needs to satisfy this equation. This is a perfectly legitimate way to define a plane. I hope this makes sense, you use two "lines" that are intuitively "different" and if you can add these vectors together to get to a different point, that point is on the plane. So you can think of a plane as a locus of points that satisfy this property. On a line, it's just a locus of points that satisfy $(x,y,z)+t\cdot \vec n$.

Now: the reason $ax+by+cz=d$ is also a great way to define a plane, is because it automatically gives you the vector perpendicular to the point. For a quick moment in time, assume that $d$ is not a fixed constant, let it be a variable. consider the following 2 points on the plane:

$(x_{1},y_{1},z_{1})$ and $(x_{2},y_{2},z_{2})$.

Since they are on the plane, they both satisfy the following equation:

$$ax_{1}+by_{1}+cz_{1}=d$$ and $$ax_{2}+by_{2}+cz_{2}=d$$

Now, by subtraction we have that: $$(ax_{1}-ax_{2})+by_{1}-by_{2}+cz_{1}-cz_{2}=0$$ or, better yet: $$[a,b,c]\cdot[x_{1}-x_{2},y_{1}-y_{2},z_{1}-z_{2}]=0$$

Since $[x_{1}-x_{2},y_{1}-y_{2},z_{1}-z_{2}]$ is a vector made by points on the plane, it is a vector on the plane. Since the dot product is zero, $[a,b,c]$ is perpendicular to this vector. Notice that this property does not depend on d. Pretty much, imagine a vector perpendicular to the floor. You can stack sheets of paper up to the roof of your house, and they will all be perpendicular to that same vector, because they are parallel to the floor. When you fix $d$ as a constant, all you do is choose one of those sheets of paper. Pretty much, it fixes it in space, it means that not only is it perpendicular to a vector, but now it is fixed in space.

By analogy: imagine $y=mx+b$. All of these lines have a slope of $m$, but you can shift $b$ up and down. That's kind of how the constant $d$ acts with a plane.

Both of these ways are pretty much the same thing, but one is looking at how two lines fully describe a plane, and the other just gives a perpendicular vector and a fixed place in space. These are both just infinite sheets.

If it helps, think about the $x,y$ plane. You can describe every point on it as a combination of $x$ and $y$, so that it looks like $(x,y)$. But really, what you're doing is looking at two different perpendicular vectors [in jargon: bases]: $(0,0,0)+t\cdot[1,0,0]+s \cdot[0,1,0]$.

Answer 2

If you consider how a line works in the xy-plane as being one variable is set by the other one, then a plane in the xyz space is where 2 variables are setting the third. For example, the xy-plane could be defined by the equation $z=0$ as one could choose any value for x and y here. That would be my intuitive way to read the equation of a plane which is usually defined as a dot product of the normal vector with $(x,y,z)$ equalling a constant.

Answer 3

A plane in $\mathbb{R}^3$ has an equation of the form $Ax+By+Cz=D$ it is hard to think about what that means when we change all three variables at once, but it is more manageable if we try to change one and see what happen. For example, starting at any point $(x,y,z)$ on the plane, if we try to increase $x$ by one, then we need to compensate for that by (assuming all positive constants) decreasing y or z. If we choose to fix one, say z, then we need to decrease y by exactly $A/B$ to have everything add up to D still. This didn't depend on which point I started at, so this "rate of change" is constant(just like a line in $\mathbb{R}^2$!) so in some sense, I could keep traveling in the direction $(1,A/B,0)$ forever and always stay in the plane. Similarly if I fixed x then y and z would have a constant rate of change and I could go forever in that direction. In fact, I don't even need to fix a variable. Starting at any point in the plane, if I head towards another point in the plane, I can travel in that direction forever while still being in the plane.

Thus, the equation corresponds to a flat infinite sheet. I hope this helps.

Answer 4

Lets take a step back.

Think about lines in 2D. Generally the equation you'd see is $y = ax +b$ but it's equivalent to $ay + bx = c$. Now, a line in 2D has no beginnings or ends it just goes off to infinity.

The same idea hold for 3D with the planes. The equation for a plane would be $ax + by + cz = d$. If z = 0 we will start off with a line. Lets now allow z to take a different value say $z_0$. The equation now reads: $ax + by = d - cz_0$, which is again a line but now we moved it down a bit. If we do it for all values of z we will get infinetly many lines stacked on top of each other creating a plane.

Since there are no bounds on z the plane goes on forever in the z direction. Since if we look at a fix z we simply look at a 2D line it will also go on forever in the x and y directions.

Thus planes aren't squares - they are unbound.

Answer 5

Another way to think of a plane in $\mathbb R^3$ is as follows. Pick any two distinct points. The set of all points in $\mathbb R^3$ which are equidistant from those two points is a plane. Every plane in $\mathbb R^3$ can be realized in this way (although the pair of points used is not uniquely determined by the plane).

Answer 6

A plane is dual to a point in 3D. Anything you can do with points planes and lines you can also do by interchanging all points for planes and all planes for points.

For example:

• Two (non-coincident) points define a unique line (join)
• Two (non-parallel) planes define a unique line (intersection)
• Two (intersecting) lines meet at a point
• Two (intersecting) lines defined a plane
• Three points define a plane
• Three planes meet at a point
• A line defines a point where it meets a plane
• A line defines a plane when it joins a point

...

The list goes on. So think of planes as the evil twins of points. Both require three coordinates to specify and in fact every (non-zero) point in $\Bbb R^3$ defines a unique plane. How? Consider a line between the origin and the point, and the plane will be perpendicular to the line through the point. Similarly, any plane defines a unique point where the line normal to the plane and through the origin intersects the plane.