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Section 4.6 Equations of Planes in \(\R^3\)

Subsection 4.6.1 Three Dimensional Point-normal Form

Suppose we start with a nonzero vector \(\vec n=(a,b,c)\text{.}\) We then consider the plane (through \(\vec0\)) that is perpendicular to \(\vec n\text{.}\) The picture below illustrates the situation. Any point \((x,y,z)\) on this plane must be perpendicular to \(\vec n\text{.}\) This means that \((x,y,z)\cdot\vec n= ax+by+cz=0\text{.}\)

Figure 4.6.1.

In addition, the converse is true, that is, if we have numbers \(a\text{,}\) \(b\) and \(c\text{,}\) not all zero, then the set of all points \((x,y,z)\) satisfying \(ax+by+cz=0\) must be perpendicular to \(\vec n=(a,b,c)\) and so is a plane through \(\vec0\text{.}\) In this case \(\vec n\) is called the normal vector to the plane.

Next we consider the equation of a plane passing through an arbitrary point \((x_0,y_0,z_0)\text{.}\) The picture in this case is pretty similar to the previous one:

Figure 4.6.3.

In this case the perpendicularity means that \((x-x_0,y-y_0,z-z_0)\cdot(a,b,c)=0\text{.}\) In other words,

\begin{equation*} a(x-x_0)+b(y-y_0)+c(z-z_0)=0 \end{equation*}

Subsection 4.6.2 General Equation of a Plane

The point-normal equation of a plane may be written as

\begin{equation*} a(x-x_0)+b(y-y_0)+c(z-z_0)=ax+by+cz-(ax_0+by_0+cz_0)=0 \end{equation*}

If we set \(d=-(ax_0+by_0+cz_0)\text{,}\) we get the general equation of a plane.

Different values \(a\text{,}\) \(b\text{,}\) \(c\text{,}\) and \(d\) may give the same plane. If we consider the planes with equations

\begin{gather*} 2x+3y+4z=0\\ 4x+6y+8z=0 \end{gather*}

then any vector \((x,y,z)\) that satisfies the first equation clearly satisfies the second one. More generally, if \(ax+by+cz+d=0\text{,}\) then for any real number \(r\not=0\) we have \(r(ax+by+cz+d)=0\) and so \((ra)x+(rb)y+(rc)z+rd=0\text{.}\) If we let \(a'=ra\text{,}\) \(b'=rb\text{,}\) \(c'=rc\) and \(d'=rd\text{,}\) then

\begin{equation*} ax+by+cz+d=0 \quad\textrm{ if and only if }\quad a'x+b'y+c'z+d'=0 \end{equation*}

and so they are the general equation of the same plane. We may write \((a',b',c',d')=r(a,b,c,d)\text{.}\) This is the exact condition we need to have two different equations of the same plane.

First, if \((a_2,b_2,c_2,d_2)=r (a_1,b_1,c_1,d_1)\text{,}\) and \((x_0,y_0,z_0)\) is in the first plane, then

\begin{equation*} a_1x_0+b_1y_0+c_1z_0+d_1=0. \end{equation*}

Multiplying both sides of the equation by \(r\) gives

\begin{align*} 0 \amp= r(a_1x_0+b_1y_0+c_1z_0+d_1)\\ \amp =ra_1x_0+rb_1y_0+rc_1z_0+rd_1\\ \amp =a_2x_0+b_2y_0+c_2z_0+d_2 \end{align*}

and so \((x_0,y_0,z_0)\) is in the second plane. A similar argument shows that any vector in the second plane is also in the first plane.

Next, suppose that the two equations are equations of the same plane. We consider this as a system of linear equations with augmented matrix

\begin{equation*} \begin{bmatrix} a_1\amp b_1\amp c_1\amp -d_1\\ a_2\amp b_2\amp c_2\amp -d_2 \end{bmatrix} \end{equation*}

which we put into reduced row echelon form to find all points in both planes. It takes just two elementary row operations. The first nonzero entry in the first row is changed to a one using \(R_1\gets \lambda_1R_1\) and the entry below this leading one is changed to a zero using \(R_2=R_2-\lambda_2R_1\text{.}\) This means that the second row in the reduced row echelon form is

\begin{equation*} (a_2,b_2,c_2,d_2)-\lambda_2\lambda_1(a_1,b_1,c_1,d_1)\text{.} \end{equation*}

However, the first and second equations together have the same solutions as the first equation alone, that is, the second equation is actually superfluous. This makes the second row all zero, and so, setting \(r=\lambda_1\lambda_2\text{,}\) we have

\begin{equation*} (a_2,b_2,c_2,d_2)=r(a_1,b_1,c_1,d_1)\text{.} \end{equation*}

Subsection 4.6.3 The equation of a plane through three given points

Suppose we have three (presumably noncollinear) points \((x_0,y_0,z_0)\text{,}\) \((x_1,y_1,z_1)\) and \((x_2,y_2,z_2)\text{.}\) We want to find the equation of the plane containing the three points. We will give three different methods, and apply each method to a specific example.

Finding the plane containing three points: method 1.

The general equation of the plane is

\begin{equation*} ax+by+cz+d=0 \end{equation*}

for some \(a\text{,}\) \(b\) and \(c\) not all zero. Substitute the three points in the equation to get three equations in four unknowns (the unknowns are \(a\text{,}\) \(b\text{,}\) \(c\) and \(d\)).

Example of this method: Let the three points be \((1,1,3)\text{,}\) \((1,0,2)\) and \((2,1,1)\text{.}\) Each point gives an equation:

\begin{align*} a+b+3c+d\amp=0 \tag{from the point \((1,1,3)\)}\\ a\phantom{+2b}+2c+d\amp=0\tag{from the point \((1,0,2)\)}\\ 2a+b+c+d\amp=0\tag{from the point \((2,1,1)\)}\text{.} \end{align*}

The augmented matrix for this system is

\begin{equation*} \left[\begin{array}{cccc|c} 1\amp1\amp3\amp1\amp0\\ 1\amp0\amp2\amp1\amp0\\ 2\amp1\amp1\amp1\amp0 \end{array}\right] \end{equation*}

which has the reduced row echelon form:

\begin{equation*} \left[\begin{array}{cccc|c} 1\amp0\amp0\amp\frac12\amp0\\ 0\amp1\amp0\amp-\frac14\amp0\\ 0\amp0\amp1\amp\frac14\amp0 \end{array}\right] \end{equation*}

This means that \(d\) is a free variable, \(c=-\frac14 d\text{,}\) \(b=\frac14 d\) and \(a=-\frac12 d\text{.}\) This makes the equation of the plane

\begin{equation*} ax+by+cz+d=-\frac12 d x+\frac14 d y-\frac14 d z +d =0 \end{equation*}

Dividing the equation by \(d\) gives the equation

\begin{equation*} -\frac12 x+\frac14 y-\frac14 z +1 =0 \end{equation*}

which is an equation of the plane. Multiplying by \(-4\) can make the equation a bit cleaner:

\begin{equation*} 2x-y+z-4=0 \end{equation*}

As a check, for \((1,1,3)\) we have \(2x-y+z-4=2-1+3-4=0\text{,}\) for \((1,0,2)\) we have \(2x-y+z-4=2-0+2-4=0\) and for \((2,1,1)\) we have \(2x-y+z-4=4-1+1-4=0\text{.}\)

Finding the plane containing three points: method 2.

The point-normal equation of a plane is

\begin{equation*} a(x-x_0)+b(y-y_0)+c(z-z_0)=0 \end{equation*}

for some \(a\text{,}\) \(b\) and \(c\) not all zero. Use one of the points as \((x_0,y_0,z_0)\) and then substitute the other two points in the equation to get two equations in three unknowns (the unknowns are \(a\text{,}\) \(b\) and \(c\)).

Example of this method: Let the three points be \((1,1,3)\text{,}\) \((1,0,2)\) and \((2,1,1)\text{.}\) Let \((x_0,y_0,z_0)=(1,1,3)\) so that the equation of the plane is

\begin{equation*} a(x-1)+b(y-1)+c(z-3)=0\text{.} \end{equation*}

Using the other two points we get

\begin{align*} -b-c \amp=0\\ a-2c \amp=0 \end{align*}

The augmented matrix of these equations is

\begin{equation*} \left[ \begin{array}{ccc|c} 0\amp-1\amp-1\amp0\\ 1\amp0\amp-2\amp0 \end{array} \right] \end{equation*}

whose reduced row echelon form is

\begin{equation*} \left[ \begin{array}{ccc|c} 1\amp0\amp-2\amp0\\ 0\amp1\amp1\amp0 \end{array} \right] \end{equation*}

which implies \(a=2c\) and \(b=-c\text{.}\) Hence

\begin{equation*} 2c(x-1)-c(y-1)+c(z-3)=0 \end{equation*}

and so

\begin{equation*} 2(x-1)-(y-1)+(z-3)=0 \end{equation*}
Finding the plane containing three points: method 3.

Once again, the point-normal equation of a plane is

\begin{equation*} a(x-x_0)+b(y-y_0)+c(z-z_0)=0 \end{equation*}

for some \(a\text{,}\) \(b\) and \(c\) not all zero. This equation says that the vectors \((a,b,c)\cdot(x-x_0,y-y_0,z-z_0)=0\) and hence they are orthogonal for any \((x,y,z)\) in the plane.

Let \((x_0,y_0,z_0)\) be one of the three points. Using each of the other two points as \((x,y,z)\text{,}\) we have two vectors \((x-x_0,y-y_0,z-z_0)\) that need to be orthogonal to \((a,b,c)\text{.}\) Since the cross product of the two vectors is orthogonal each of them, that cross product may be used as the vector \((a,b,c)\text{.}\)

Example of this method.

Let the three points be \((1,1,3)\text{,}\) \((1,0,2)\) and \((2,1,1)\text{.}\) Let \((x_0,y_0,z_0)=(1,1,3)\text{.}\) Then the other two points give \((x-x_0,y-y_0,z-z_0)\) as \((0,-1,-1)\) and \((1,0,-2)\text{.}\) It's easy to compute \((0,-1,-1)\times(1,0,-2)=(-2,-1,1)\text{,}\) and so we may use \((a,b,c)=(2,-1,1)\text{.}\) In other words, the equation of the plane is

\begin{equation*} 2(x-1)-(y-1)+(z-3)=0 \end{equation*}

If we expand this final answer, we get

\begin{equation*} 2x-y+z-4=0 \end{equation*}

All three methods give the right answer, but clearly the third one is the easiest to use. It's the theory behind the cross product that enables a much easier computation. This happens often in many areas of mathematics: knowing some theory can make your life much easier.

Here is the graph of the plane with equation \(2x-y+z-4=0\) in \(\R^3\) including the three points \((1,1,3)\text{,}\) \((1,0,2)\) and \((2,1,1)\) used to determine it. The plane is blue, and the line joining the origin to \((a,b,c)=(2,-1,1)\) is red. As expected with the point-normal form, the line is orthogonal to the plane.

Figure 4.6.7.

Subsection 4.6.4 The area of a triangle and parallelogram determined by three points

Suppose we have three points \(A\text{,}\) \(B\) and \(C\) in \(\R^3\text{.}\) We want to compute the area of the triangle determined by these points. If the points are collinear, then the triangle collapses to a line and the area is zero, so we may suppose that the points are noncollinear.

Figure 4.6.8.

We define two vectors:

\begin{equation*} \vec u= \overrightarrow{AB}\\ \vec v=\overrightarrow{AC} \end{equation*}

The area of the triangle is \(\frac12 hb\) where \(b=\|\vec u\|\) and \(h=\|\vec v\|\sin(\theta)\text{.}\) Hence the area of the triangle is \(\frac12 \|\vec u\|\|\vec v\|\sin(\theta) = \frac12\|\vec u\times\vec v\|\text{.}\)

The area of parallelogram follows easily from the triangle (and vice-versa) since the area of the parallelogram is twice that of the triangle.

Figure 4.6.9.

The area of the parallelogram is \(\|\vec u\times\vec v\|\text{.}\)