Plane parametrization example

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

x - 2y + 3z = 18. (1)

A parametrization for a plane can be written as

x = sa + tb + c    

where a and b are vectors parallel to the plane and c is a point on the plane. The parameters s and t are real numbers. Any point x on the plane is given by sa + tb + c for some value of (s, t). Any value of (s, t) corresponds to a point x on the plane.

Note that x is different from x; in fact, x is the vector 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 (1) means that

z = $\displaystyle {\frac{{18-x+2y}}{{3}}}$ = $\displaystyle {\frac{{18-0+2(0)}}{{3}}}$ = 6.    

A point on the plane is 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 n = (1, - 2, 3). (Why?) So, we need to find two vectors a and b that are perpendicular to n, i.e., we need a . n = 0 and b . n = 0. If a = (a1, a2, a3) and b = (b1, b2, b3), we need

a1 -2a2 +3a3 = 0 (2)

and

b1 -2b2 +3b3 = 0 (3)

There are many choices for a and b. In fact, we can choose any two components of a or b and use the above condition to specify the third component. To keep life simple, we set a1 = 1 and a2 = 0. Then, by equation (2), we know that

a3 = $\displaystyle {\frac{{-a_1+2a_2}}{{3}}}$ = - $\displaystyle {\frac{{1}}{{3}}}$.    

We conclude that the vector a = (1, 0, - 1/3) is parallel to the plane.

We need to choose b so that b is not parallel to a. To ensure this, we set b1 = 0 and b2 = 1. Then, by equation (3), we know that

b3 = $\displaystyle {\frac{{-b_1+2b_2}}{{3}}}$ = $\displaystyle {\frac{{2}}{{3}}}$.    

We conclude that the vector b = (0, 1, 2/3) is parallel to the plane. (You should double check this for both a and b.)

We are finished. A parametrization for the plane is

x = s(1, 0, - 1/3) + t(0, 1, 2/3) + (0, 0, 6) = (s, t, 6 - s/3 + 2t/3).    

Since x = (x, y, z), we could write this as

\begin{displaymath}\begin{cases}
x=s\\ y=t\\ z=6-s/3+2t/3. \end{cases}\end{displaymath}    

(If we had made different choices for a1, a2, b1, and b2, 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 a1 = 1, a2 = 0, b1 = 0, and b2 = 1 and chose the first two components of 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

z = $\displaystyle {\frac{{18-x+2y}}{{3}}}$    

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

z = $\displaystyle {\frac{{18-s+2t}}{{3}}}$ = 6 - s/3 + 2t/3.    

Again, we end up with the parametrization

\begin{displaymath}\begin{cases}
x=s\\ y=t\\ z=6-s/3+2t/3. \end{cases}\end{displaymath}    

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



Duane Nykamp
nykamp@math.umn.edu
2005-01-25