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

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

$${\frac{{18-x+2y}}{{3}}} {\frac{{18-0+2(0)}}{{3}}}$$

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

$${\frac{{-a_1+2a_2}}{{3}}} {\frac{{1}}{{3}}}$$

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

$${\frac{{-b_1+2b_2}}{{3}}} {\frac{{2}}{{3}}}$$

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

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

(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

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

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

$${\frac{{18-s+2t}}{{3}}}$$

Again, we end up with the parametrization

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

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