Definition 1.3.1.
Consider an equation in the variables \(x,y,z\text{.}\) The locus in \(\RR^3\) of this equation is the set of vectors that satisfy the equation. The plural of a locus is loci.
Section 1.3 Equations of Planes
Subsection 1.3.1 Definition of a Locus
There are many ways to mathematically describe an geometric object. Cartesian geometry allows for perhaps the most powerful idea: the locus. I’ll define this in \(\RR^3\text{,}\) but the definition is general to any Euclidean space.
In general, the equation can be of any sort. The unit circle in \(\RR^2\) is most commonly defined as the locus of the equation \(x^2 + y^2 = 1\text{.}\) The graph of a single-variable function is the locus (also in \(\RR^2\)) of the equation \(y = f(x)\text{.}\) My goal in this section is equation of places, which are flat. Flat objects arise as loci of linear equations. Let me remind you of that definition.
Definition 1.3.2.
Let \(a\text{,}\) \(b\text{,}\) \(c\) and \(d\) be real numbers. A linear equation in the variables \(x,y,z\) is an equation of the following form.
\begin{equation*}
a x + b y + c z = d
\end{equation*}
The most familiar linear loci are lines. Very briefly, let me redefined a line in \(\RR^2\text{.}\) (A line in \(\RR^3\) needs a different definition; in this course, one way to describe lines is as parametric curves, as in Section 2.1.)
Definition 1.3.3.
A line in \(\RR^2\) is the locus of the equation \(ax + by = c\) for \(a,b,c \in \RR\text{.}\) The line goes through the origin if \(c=0\text{.}\)
Definition 1.3.4.
A plane in \(\RR^3\) is the locus of the linear equation \(ax + by + cz = d\text{.}\) The plane goes through the origin if \(d=0\text{.}\)
Notes that the equation for a locus is not unique; many equations may describe the same locus. Since they are equations, I can change them by doing the same thing to both sides (hence preserving the equality). Consider the plane given by this equation.
\begin{equation*}
3x - y + 5z = 12
\end{equation*}
I can multiply both sides by \(-1\) and still describe the same plane.
\begin{equation*}
-3x + y - 5z = -12
\end{equation*}
I could also multiply the original equation by any constant, say \(100\text{,}\) and still describe the same plane.
\begin{equation*}
300x - 100y + 500z = 1200
\end{equation*}
Finally, I must be careful multiplying by zero (or any expressing that might evaluate to zero). If I multiply both sides of the equation by zero, I get a trivial equation.
\begin{equation*}
(0)(3x - y + 5z) = (0)(12) \implies 0 = 0
\end{equation*}
The equation \(0=0\) describes all points, not just the plane, so I have destroyed the locus. As with many other places in mathematics, multiplying or dividing by zero always needs to be the back of you mind as something to be careful about.
Subsection 1.3.2 Dot Products and Loci
Loci are the solutions to one of more linear equations. These linear equation can be re-interpreted and expressed via via dot products. Consider, again, the linear equation in \(\RR^3\text{.}\)
\begin{equation*}
ax + by + cz = d
\end{equation*}
I can think of the variables \(a,b,c\) as the components of a vector \(u = (a,b,c) \in \RR^3\text{.}\) Let me use \(v\) to also label the vector of variables: \(v = (x,y,z)\text{.}\) I can re-write the linear equation using these two vectors.
\begin{align*}
ax + by + cd \amp = d\\
(a,b,c) \cdot (x,y,z) \amp = d \\
u \cdot v \amp = d
\end{align*}
In this way, a linear equation specifies that the dot product result of a variable vector \(v\) with a fixed vector \(u\) must have the result \(d\text{.}\) In this light, a plane in \(\RR^3\) is given by the equation
\begin{equation*}
(a,b,c) \cdot (x,y,z) = d\text{.}
\end{equation*}
This plane is precisely all vectors whose dot product with the vector \((a,b,c)\) produces the fixed number \(d\text{.}\) If \(d=0\) (i.e., if the plane goes through the origin), the equation become
\begin{equation*}
(a,b,c) \cdot (x,y,z) = d\text{.}
\end{equation*}
A dot product of zero indicates that the two inputs to the dot product are perpendicular. Therefore, a plane through the origin is the set of all vectors which are perpendicular to a fixed vector \((a,b,c)\text{.}\) This leads to an important definition.
Definition 1.3.5.
Let \(P\) be a plane in \(\RR^3\) determined by the equation
\begin{equation*}
(a,b,c) \cdot (x,y,z) = d \text{.}
\end{equation*}
The vector \((a,b,c)\) is called the normal to the plane.
If \(d=0\text{,}\) the plane or hyperplane is perpendicular to its normal. The remark thing is this: the notion of orthogonality of the normal still works when \(d \neq 0\text{.}\) In this case, the normal is a local perpendicular direction from any point on the plane. Treating any such point as a local origin, the normal points in a direction perpendicular to all the local direction vectors which lie on the plane. This is the key of understanding and describing planes in \(\RR^3\text{.}\)
Subsection 1.3.3 An Algorithm for Equations of Planes
Now I can build a general process for finding the equation of a plane in \(\RR^3\text{.}\) Using the equations above, I need two pieces of information: the (local direction) normal \((a,b,c)\) and the constant \(d\text{.}\)
I’m going to describe three ways of calcualting the equation of a plane in \(\RR^3\text{,}\) each starting from different initial information. All three of these collections of data uniquely determine a plane.
- A point and the normal.
- A point and two local directions.
- Three points.
I’ll describe the algorithm for all three of these in order, since the calculations get longer in each step of this list.
In the first description, I am actually given the normal \((a,b,c)\) and some point \((x_0, y_0, z_0)\) on the plane, I only need to calculate \(d\text{.}\) To do so, I just evaluate the left side of the equation of the plane: doing so is equal to the right side, which is just \(d\text{.}\) Let me show some examples.
Example 1.3.6.
Say that I am given the normal \((0, -2, -3)\) and the point \((1,-7,5)\text{.}\) What is the unique plane described by this data?
I calculate the left side of the equation of the plane, which is the dot product of the point and the normal. This will result in the constant \(d\text{.}\)
\begin{equation*}
(0,-2,-3) \cdot (1, -7, 5) = 0 + 14 - 15 = -1 = d
\end{equation*}
So \(d = -1\text{,}\) which means I can write the equation of the plane. The coefficients of the variables are the coordinates of the normal.
\begin{equation*}
0x - 2y - 3z = -1
\end{equation*}
There is always some variability in how to write these equations. Since the \(x\) variable is multiplied by zero, I can remove it. I can also multiply both sides by \(-1\) if I wanted to.
\begin{equation*}
2y + 3z = 1
\end{equation*}
This equation also describes the same plane.
Example 1.3.7.
Say that I am given the normal \((-5, 1, 4)\) and the point \((0,3,11)\text{.}\) What is the unique plane described by this data?
I calculate the left side of the equation of the plane, which is the dot product of the point and the normal. This will result in the constant \(d\text{.}\)
\begin{equation*}
(-5,1,4) \cdot (0,3,11) = 0 + 3 + 44 = 47
\end{equation*}
So \(d = 47\text{,}\) which means I can write the equation of the plane. The coefficients of the variables are the coordinates of the normal.
\begin{equation*}
-5x + y + 4z = 47
\end{equation*}
Example 1.3.8.
Say that I am given the normal \(\left( \frac{-4}{7},
\frac{-11}{13}, \frac{5}{2} \right)\) and the point \(\left( \frac{1}{4}, \frac{-1}{2}, \frac{-4}{13}
\right)\text{.}\) What is the unique plane described by this data?
I calculate the left side of the equation of the plane, which is the dot product of the point and the normal. This will result in the constant \(d\text{.}\)
\begin{equation*}
\left( \frac{-4}{7},\frac{-11}{13}, \frac{5}{2} \right)
\cdot \left( \frac{1}{4}, \frac{-1}{2}, \frac{-4}{13}
\right) = \frac{-4}{28} + \frac{11}{26} - \frac{20}{26} =
\frac{-89}{182}
\end{equation*}
So \(d = -\frac{89}{182}\text{,}\) which means I can write the equation of the plane. The coefficients of the variables are the coordinates of the normal.
\begin{equation*}
\frac{-4}{7} x - \frac{11}{13} y + \frac{5}{2} z =
\frac{-89}{182}
\end{equation*}
Again, there is variability in how to write the equation. If I wanted to clear denominators, I could multiply through by \(182\text{.}\)
\begin{equation*}
-104x - 154y + 455 z = -89
\end{equation*}
This equation also describes the same plane.
If I am given a point on a plane and two local direction on the plane, then I need to construct the normal first. The normal is perpendicular to any local direction on the plane. This is where the cross product comes is: the cross product is a operation which takes two vectors and produces a third vector perpendicular to the first two. Therefore, the cross product of the two local directions is the normal. After I have the normal, I can proceed as before, using the equation and the point to calculate the constant \(d\text{.}\) Let me do one example.
Example 1.3.9.
Say that I am given the point \((0, 3, 7)\) and the local directions \((-1, -5, 3)\) and \((-5, 0, 1)\text{.}\) What is the equation of the plane described by this data?
I take the cross product of the two local directions to ge the normal to the plane.
\begin{equation*}
(-1,-5,3) \times (-5,0,1) = (-5 - 0, -15 - (-1), 0 - 25) =
(-5, -14, -25)
\end{equation*}
Then I proceed as before. I calculate the left side of the equation of the plane, which is the dot product of the point and the normal. This will result in the constant \(d\text{.}\)
\begin{equation*}
(-5, -14, -25) \cdot (0,3,7) = -42 - 175 = -217
\end{equation*}
So \(d = -217\text{,}\) which means I can write the equation of the plane. The coefficients of the variables are the coordinates of the normal.
\begin{equation*}
-5x - 14y - 25 z = -217
\end{equation*}
I could multiply through by \(-1\) to make everything positive, if I wanted.
\begin{equation*}
5x + 14y + 25 z = 217
\end{equation*}
This equation also describes the same plane.
Finally, a plane can be determined by three points (as long as the three points do not all lie on the same line). If I am given three points (\(p\text{,}\) \(q\) and \(r\)), then I can use \(p\) as the local origin and construct the local direction vectors as \(q-p\) and \(r-p\text{.}\) Then I can proceed as in the previous step: the normal is the cross product of the two local direction, \((q-p)
\times (r-p)\text{.}\) Finally, I finish with the first step, using the normal and a point to find the constant \(d\text{.}\) Note, with this data, I have three points to chose from for this final step. All three should work and produce the same constant. If they fail to do so, that means there is an error earlier in the work somewhere. I’ll do an example to show all the steps.
Example 1.3.10.
What is the equation of the plane determined by the three points \((0,4,0)\text{,}\) \((-3, -6, 2)\) and \((-7, 0,
1)\text{.}\)
First I need to calculate the local direction. I’ll set \((0,4,0)\) as my local origin. Then I calculate the differences of the two other points from this point.
\begin{gather*}
(-3, -6. 2) = (0,4,0) = (-3, -10, 2) \\
(-7, 0. 1) = (0,4,0) = (-7, -4, 1)
\end{gather*}
These are the two local direction. Then I proceed as with the previous case. I start by taking the cross product of the two local directions to ge the normal to the plane.
\begin{align*}
(-3, -10, 2) \times (-7, -4, 1) \amp = (-10 - (-8),
(-14, - (-3), 12 - 70))\\
\amp = (-2, -11, -58)
\end{align*}
Then I again proceed as before, going back to the very first case. I calculate the left side of the equation of the plane, which is the dot product of the point and the normal. This will result in the constant \(d\text{.}\) I use the first point, but I could use any of the three points.
\begin{equation*}
(-2, -11, -58) \cdot (0,4,) = 0 - 44 + 0 = -44
\end{equation*}
So \(d = -44\text{,}\) which means I can write the equation of the plane. The coefficients of the variables are the coordinates of the normal.
\begin{equation*}
-2x - 11y - 58 z = -44
\end{equation*}
I could multiply through by \(-1\) to make everything positive, if I wanted.
\begin{equation*}
2x + 11y + 58z = 44
\end{equation*}
This equation also describes the same plane.
