I have the vector form of a line:
$\begin{bmatrix}x\\y\\z\end{bmatrix} =
\begin{bmatrix}1\\0\\-2\end{bmatrix} +
t\begin{bmatrix}3\\-2\\5\end{bmatrix}$ where $t \in \mathbb{R}$
Normally, to convert a vector equation for a plane in $\mathbb{R}^3$ to a general equation, I can simply solve the parametric equations associated with the vector equation.
What is happening when I solve the parametric equations of the vector equation for a line in the following manner?
$x=1+3t\\y=-2t\\z=-2+5t$
$\Rightarrow z+y-x = -3$
I now have an equation for a plane? My guesses:
- Is this one of many possible solutions?
- Are there infinite solutions to this system of equations, with each one being a plane containing the line?
Best Answer
In $$\Rightarrow z+y-x = -3$$
You have not solved anything. What you have done is to eliminate $t$ through addition. This plane is one of infinitely planes on which the stated line lies.
Doing things differently, for example, I can get:
$z-2y-3x = -5$
by multiplying the second equation by $-2$ and the first one by $-3$, then summing everything.