Skip to main content

Linear Algebra For Undergraduate Students

Yilan(Peter) Tan

Section 8.1 9.1 Planes in Space and Their Equations

Subsection 8.1.1 Point-Normal Form of a Plane

Definition 8.1.1. Point-Normal Equation of a Plane.

Let \(P_0(x_0,y_0,z_0)\) be a fixed point in space and let \(\mathbf{n}=(A,B,C)\neq \mathbf{0}\) be a normal vector. The plane through \(P_0\) with normal vector \(\mathbf{n}\) is given by
\begin{equation*} A(x-x_0)+B(y-y_0)+C(z-z_0)=0. \end{equation*}
🔗
🔗
described in detail following the image
A plane in space with a marked point on the plane and a normal vector perpendicular to the plane.
🔗
Figure 8.1.2. Geometric Meaning of the Point-Normal Equation
🔗

Proof.

By geometry, \(P\) is on the plane exactly when the displacement vector
\begin{equation*} \overrightarrow{P_0P}=(x-x_0,\,y-y_0,\,z-z_0) \end{equation*}
is perpendicular to the normal vector \(\mathbf{n}\text{.}\) By the dot-product criterion for orthogonality,
\begin{equation*} \mathbf{n}\cdot \overrightarrow{P_0P}=0. \end{equation*}
Therefore,
\begin{equation*} (A,B,C)\cdot(x-x_0,\,y-y_0,\,z-z_0)=0, \end{equation*}
which is exactly
\begin{equation*} A(x-x_0)+B(y-y_0)+C(z-z_0)=0. \end{equation*}
🔗
🔗

Example 8.1.4. Building a Plane Equation from a Point and a Normal Vector.

Find the equation of the plane through \(P_0(1,-2,3)\) with normal vector \(\mathbf{n}=(2,-1,4)\text{.}\)
🔗
Solution.
By Definition 8.1.1,
\begin{equation*} 2(x-1)-1(y+2)+4(z-3)=0. \end{equation*}
Simplifying,
\begin{equation*} 2x-y+4z-16=0. \end{equation*}
🔗
🔗
🔗
🔗

Subsection 8.1.2 General Form and Geometric Meaning of Coefficients

Definition 8.1.5. General Form of a Plane.

Every plane in space can be written as
\begin{equation*} Ax+By+Cz+D=0, \end{equation*}
where \((A,B,C)\neq (0,0,0)\text{.}\) The vector \((A,B,C)\) is a normal vector to the plane.
🔗
🔗

Insight 8.1.6. What Do the Coefficients Mean?

In \(Ax+By+Cz+D=0\text{,}\) the coefficients \(A,B,C\) determine the direction of the normal vector, so they determine the orientation of the plane. The constant term \(D\) shifts the plane along the normal direction.
🔗
If two planes have proportional triples \((A,B,C)\text{,}\) then their normals are parallel, so the planes are parallel or identical.
🔗
🔗
described in detail following the image
Two parallel planes in space with aligned normal vectors to highlight coefficient proportionality in their equations.
🔗
Figure 8.1.7. Parallel Planes Share Parallel Normal Vectors
🔗
🔗

Subsection 8.1.3 Intercept Form of a Plane

Proof.

The intercept points are \(A(a,0,0)\text{,}\) \(B(0,b,0)\text{,}\) and \(C(0,0,c)\text{.}\) Substituting each into a general linear equation shows that the coefficients can be scaled to obtain
\begin{equation*} \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1. \end{equation*}
🔗
🔗
described in detail following the image
A plane cutting the x-, y-, and z-axes at three intercept points, visualizing the intercept form equation.
🔗
Figure 8.1.9. Plane Determined by Three Axis Intercepts
🔗

Example 8.1.10. Converting General Form to Intercept Form.

Convert \(2x+3y+6z-12=0\) to intercept form and identify its axis intercepts.
🔗
Solution.
Rewrite as \(2x+3y+6z=12\text{,}\) then divide by \(12\text{:}\)
\begin{equation*} \frac{x}{6}+\frac{y}{4}+\frac{z}{2}=1. \end{equation*}
Hence the intercepts are \((6,0,0)\text{,}\) \((0,4,0)\text{,}\) and \((0,0,2)\text{.}\)
🔗
🔗
🔗

Checkpoint 8.1.11. Practice Exercises.

(a)
Find the plane through \((2,1,-1)\) with normal vector \((3,-2,5)\text{.}\)
🔗
🔗
(b)
Determine whether the planes \(2x-y+z-3=0\) and \(4x-2y+2z+1=0\) are parallel or identical.
🔗
🔗
(c)
Find the intercept form of \(3x+2y+z=6\text{.}\)
🔗
🔗
🔗
🔗
🔗
PrevTopNext