Multivariable calculus and vector analysis

Plane parametrization example

Example: Find a parametrization of (or a set of parametric equations for) the plane

$$\begin{array}{lll}\hfill x-2y+3z=18.& & \hfill \text{ (1)}\end{array}$$

A parametrization for a plane can be written as

$$\begin{array}{lll}\hfill \mathbf{x}=s\mathbf{a}+t\mathbf{b}+\mathbf{c}& & \hfill \end{array}$$

where $\mathbf{a}$ and $\mathbf{b}$ are vectors parallel to the plane and $\mathbf{c}$ is a point on the plane. The parameters $s$ and $t$ are real numbers. Any point $\mathbf{x}$ on the plane is given by $s\mathbf{a}+t\mathbf{b}+\mathbf{c}$ for some value of $\left(s,t\right)$. Any value of $\left(s,t\right)$ corresponds to a point $\mathbf{x}$ on the plane.

Note that $\mathbf{x}$ is different from $x$; in fact, $\mathbf{x}$ is the vector $\mathbf{x}=\left(x,y,z\right)$.

Solution method 1

To find a parametrization, we need to find two vectors parallel to the plane and a point on the plane. Finding a point on the plane is easy. We can choose any value for $x$ and $y$ and calculate $z$ from the equation for the plane. Let $x=0$ and $y=0$, then equation (1) means that

$$\begin{array}{lll}\hfill z=\frac{18-x+2y}{3}=\frac{18-0+2\left(0\right)}{3}=6.& & \hfill \end{array}$$

A point on the plane is $\mathbf{c}=\left(0,0,6\right)$. (Clearly, there are many other choices.)

Now, we have to find two vectors parallel to the plane. A normal vector of the plane is $\mathbf{n}=\left(1,-2,3\right)$. (Why?) So, we need to find two vectors $\mathbf{a}$ and $\mathbf{b}$ that are perpendicular to $\mathbf{n}$, i.e., we need $\mathbf{a}\cdot \mathbf{n}=0$ and $\mathbf{b}\cdot \mathbf{n}=0$. If $\mathbf{a}=\left(a_1,a_2,a_3\right)$ and $\mathbf{b}=\left(b_1,b_2,b_3\right)$, we need

$$\begin{array}{lll}\hfill a_1-2a_2+3a_3=0& & \hfill \text{ (2)}\end{array}$$

and

$$\begin{array}{lll}\hfill b_1-2b_2+3b_3=0& & \hfill \text{ (3)}\end{array}$$

There are many choices for $\mathbf{a}$ and $\mathbf{b}$. In fact, we can choose any two components of $\mathbf{a}$ or $\mathbf{b}$ and use the above condition to specify the third component. To keep life simple, we set $a_1=1$ and $a_2=0$. Then, by equation (2), we know that

$$\begin{array}{lll}\hfill a_3=\frac{-a_1+2a_2}{3}=-\frac{1}{3}.& & \hfill \end{array}$$

We conclude that the vector $\mathbf{a}=\left(1,0,-1∕3\right)$ is parallel to the plane.

We need to choose $\mathbf{b}$ so that $\mathbf{b}$ is not parallel to $\mathbf{a}$. To ensure this, we set $b_1=0$ and $b_2=1$. Then, by equation (3), we know that

$$\begin{array}{lll}\hfill b_3=\frac{-b_1+2b_2}{3}=\frac{2}{3}.& & \hfill \end{array}$$

We conclude that the vector $\mathbf{b}=\left(0,1,2∕3\right)$ is parallel to the plane. (You should double check this for both $\mathbf{a}$ and $\mathbf{b}$.)

We are finished. A parametrization for the plane is

$$\begin{array}{llll}\hfill \mathbf{x}& =s\left(1,0,-1∕3\right)+t\left(0,1,2∕3\right)+\left(0,0,6\right)=\left(s,t,6-s∕3+2t∕3\right).& \hfill & \end{array}$$

Since $\mathbf{x}=\left(x,y,z\right)$, we could write this as

$$\begin{array}{lll}\hfill \left\{\begin{array}{c}x=s\hfill \\ y=t\hfill \\ z=6-s∕3+2t∕3.\hfill \end{array}\right.& & \hfill \end{array}$$

(If we had made different choices for $a_1$, $a_2$, $b_1$, and $b_2$, we would have come up with a different parametrization. Or, if we chose a different point, we would have come up with a different parametriaztion.)

Shortcut method

There is a quick way to come up with the particular parametrization I made above. Because I chose $a_1=1$, $a_2=0$, $b_1=0$, and $b_2=1$ and chose the first two components of $\mathbf{c}$ to be zero, we ended up wtih a parametrization where $x=s$ and $y=t$. (Can you see why this is true?)

Since equation (1) can be written

$$\begin{array}{lll}\hfill z=\frac{18-x+2y}{3}& & \hfill \end{array}$$

we can plug in those values of $x$ and $y$ to determine that

$$\begin{array}{lll}\hfill z=\frac{18-s+2t}{3}=6-s∕3+2t∕3.& & \hfill \end{array}$$

Again, we end up with the parametrization

$$\begin{array}{lll}\hfill \left\{\begin{array}{c}x=s\hfill \\ y=t\hfill \\ z=6-s∕3+2t∕3.\hfill \end{array}\right.& & \hfill \end{array}$$

Question: will the shortcut method always work? (I.e., is there a reason to know the longer method?)

Contents by topic

1. Geometry in higher dimensions

1a. Preliminaries

1b. Lines and planes

• Parametrization of a line
• Forming planes
• A line or a plane or a point?
• Parametrization of a plane
• Distance from point to plane
• Parametrization of a line examples
• Forming plane examples
• Plane parametrization example
• Distance from point to plane example
• Intersecting planes example

1c. Matrices

1d. Visualizing functions in higher dimensions

2. Differential Calculus

3. Integral calculus

4. Vector calculus

5. The fundamental theorems of vector calculus

6. Review material