Finding an equation and parametric description given 3 points

Let m be the plane through (0,1,1), (0,1,0) and (-2,-1,-1).

This concept has always confused me: How would I find the equation and parametric description given just these points??

I think the parametric description is just (0,0,1)+t(0,0,1)+s(-2,-2,1) for some t and s; but how do you derive a formula for the plane given this information?

-
A parametric equation for l, and a parametric equation for m? – J. M. Oct 16 '11 at 0:39
What do you mean by "a formula for the plane"? A parametric description is a formula for the plane. Your parametric description seems to be wrong, since the point $(0,0,1)$ that it yields isn't on the plane. Also, why did you introduce the line $l$? It doesn't occur anywhere afterwards. – joriki Oct 16 '11 at 0:42
whoops! no l, just given those 3 points for m. – khchan Oct 16 '11 at 0:42
from what I understand, I need define v=(0,1,1)-(0,1,0) = (0,0,1) and w = (-2,-1,-1)-(0,1,0)=(-2,-2,1), where v and w are vectors that span m, then the parametric form should be (0,1,0)+t(0,0,1)+s(-2,-2,1). The part where I get confused is how I represent that in the form Ax+By+Cz=D – khchan Oct 16 '11 at 0:47

The plane is parallel to both $\langle -2,-1,-1 \rangle - \langle 0,1,0 \rangle = \langle -2,-2,-1 \rangle$ and $\langle 0,1,1 \rangle - \langle 0,1,0 \rangle = \langle 0,0,1 \rangle$. The plane passes through the point $\langle 0,1,0 \rangle$ so a parametrization for the plane is ${\bf r}(s,t)= \langle 0,1,0 \rangle + s\langle -2,-2,-1 \rangle + t\langle 0,0,1 \rangle$. You can think of this as standing at the point $\langle 0,1,0 \rangle$ and then moving any amount in either $\langle -2,-2,-1 \rangle$ or $\langle 0,0,1 \rangle$ direction to get around on the plane.

To find the scalar equation for the plane you need a point and a normal vector (a vector perpendicular to the plane). You already have a point (in fact you have 3!), so you just need the normal. You've already constructed 2 vectors which are parallel to the plane so computing their cross product will give you a vector perpendicular to the plane.

$$\langle -2,-2,-1 \rangle \times \langle 0,0,1 \rangle = \begin{vmatrix} {\bf i} & {\bf j} & {\bf k} \\ -2 & -2 & -1 \\ 0 & 0 & 1 \end{vmatrix} = \langle -2,2,0 \rangle$$.

Using the normal vector $\langle -2,2,0 \rangle$ and the point $\langle 0,1,0\rangle$, the scalar equation of the plane is $(-2)(x-0)+(2)(y-1)+(0)(z-0)=0$.

Remember the parametrization is "point plus 2 parameters and 2 parallel vectors" the scalar equation is "point plus 1 vector perpendicular to the plane"

-
Thanks! I was going in completely the wrong direction. – khchan Oct 16 '11 at 1:18