Trouble expressing general form of the plane equation

Question

I have a task to convert a vector form of the plane equation to the general form. General form being $ax + by + cz + d = 0$.

This is the equation:

$\vec r = (2, 1, 3) + u(1, 0, 0) + v(-1, 1, 2)$

Now, the way were were thought was this: Express $x, y$ and $z$; express $u$ and $v$ in terms of $x, y$ and $z$ and then just put it all in one equation.

Here's the parametric form:

$x = 2 + u - v$

$y = 1 + v$

$z = 3 + 2v$

I tried to express $v$ in terms of $y$ and $z$, then I expressed $y$ in terms of $z$, but in the end I got a contradictory system. Namely, I got that $2 = 3$.

Can someone solve this step-by-step? It appears I got lost.

Answer 1

We know that $<1,0,0>$ and $<-1,1,2>$ lie in the plane.

So now we can find a vector normal to plane, which would be:

$$\begin{align}\vec n &=\ <1,0,0>\times<-1,1,2>\\ &=\ <0,-2,1> \end{align}$$

Now you have the equation as: $$(\vec r -<2,1,3>)\cdot \vec n = 0$$ $$(<x,y,z>-<2,1,3>)\cdot \vec n = 0$$

Solving yields:

$$2y-z+1=0$$

Answer 2

You need to eliminate $u$ and $v$, leaving you with an equation in $x$, $y$ and $z$.

$x = 2 + u - v$

$y = 1 + v$

$z = 3 + 2v$

Rewrite $v=y-1$ and substitute that into both the other equations:

$x = 2 + u - v \Rightarrow x=2+u-(1-y)\Rightarrow x-y-1=u$. In this case you can't eliminate $u$ but that isn't a problem because:

$z = 3 + 2v \Rightarrow z=3+2y-2 \Rightarrow 2y-z+1=0$ gives you the equation in the required format.

Note that $a=0$, $b=2$, $c=-1$, $d=1$