# Math Insight

### Plane parametrization example

Example: Find a parametrization of (or a set of parametric equations for) the plane \begin{align} x-2 y + 3z = 18. \label{planeeqn} \end{align}

A parametrization for a plane can be written as \begin{align*} \vc{x} = s \vc{a} + t \vc{b} + \vc{c} \end{align*} where $\vc{a}$ and $\vc{b}$ are vectors parallel to the plane and $\vc{c}$ is a point on the plane. The parameters $s$ and $t$ are real numbers. Any point $\vc{x}$ on the plane is given by $s \vc{a} + t \vc{b} + \vc{c}$ for some value of $(s,t)$. Any value of $(s,t)$ corresponds to a point $\vc{x}$ on the plane.

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

#### 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 \eqref{planeeqn} means that \begin{align*} z = \frac{18-x+2y}{3} = \frac{18-0+2(0)}{3} = 6. \end{align*} A point on the plane is $\vc{c} = (0,0,6)$. (Clearly, there are many other choices.)

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

There are many choices for $\vc{a}$ and $\vc{b}$. In fact, we can choose any two components of $\vc{a}$ or $\vc{b}$ 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 \eqref{aperp}, we know that \begin{align*} a_3 = \frac{-a_1+2a_2}{3} = -\frac{1}{3}. \end{align*} We conclude that the vector $\vc{a}=(1,0,-1/3)$ is parallel to the plane.

We need to choose $\vc{b}$ so that $\vc{b}$ is not parallel to $\vc{a}$. To ensure this, we set $b_1=0$ and $b_2=1$. Then, by equation \eqref{bperp}, we know that \begin{align*} b_3=\frac{-b_1+2b_2}{3} = \frac{2}{3}. \end{align*} We conclude that the vector $\vc{b}=(0,1,2/3)$ is parallel to the plane. (You should double check this for both $\vc{a}$ and $\vc{b}$.)

We are finished. A parametrization for the plane is \begin{align*} \vc{x} &= s (1,0,-1/3) + t (0,1,2/3) + (0,0,6) = (s,t,6-s/3+2t/3). \end{align*} Since $\vc{x}=(x,y,z)$, we could write this as \begin{align*} \begin{cases} x=s\\y=t\\z=6-s/3+2t/3. \end{cases} \end{align*} (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 $\vc{c}$ to be zero, we ended up with a parametrization where $x=s$ and $y=t$. (Can you see why this is true?)

Since equation \eqref{planeeqn} can be written \begin{align*} z = \frac{18-x+2y}{3} \end{align*} we can plug in those values of $x$ and $y$ to determine that \begin{align*} z= \frac{18-s+2t}{3} = 6-s/3+2t/3. \end{align*} Again, we end up with the parametrization \begin{align*} \begin{cases} x=s\\y=t\\z=6-s/3+2t/3. \end{cases} \end{align*}

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