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 = 18 - x + 2y ------------ 3 = 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 = - a1-+-2a2 3 = -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 = - b1-+-2b2 3 = 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
( |{ x = s |( y = t z = 6 - s∕3 + 2t∕3.
(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 = 18---x-+-2y- 3
we can plug in those values of x and y to determine that
z = 18---s-+-2t 3 = 6 - s∕3 + 2t∕3.
Again, we end up with the parametrization
( |{ x = s |( y = t z = 6 - s∕3 + 2t∕3.

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