The parametric form of a line

Question

For the parametric representation of a line L with the following points, is my answer correct:

P1 = <2,2,0>, P2 = <0,-2,-4>, P3 = <3,4,2>

Is this correct: X = P1 + s.P1P2 + t.P1P3 = <2,2,0> + s.<-2,-4,-4> + t.<1,2,2>?

Answer 1

Notice that you only need two points to determine a line. There is a single line (with many possible parametrisations) passing through $P_1$ and $P_2$. It may, or may not, pass through $P_3$. Let's see.

There's a really easy way two parametrise a line through to points. The line through $P_1$ and $P_2$ is parametrised by $\ell(t) := (1-t)P_1 + tP_2$. When $t=0$ you're at $P_1$ and when $t=1$ you're at $P_2$.

\begin{array}{ccc} \ell(t) &=& (1-t)P_1 + tP_2 \\ &=& (1-t)\langle 2,2,0 \rangle + t \langle 0, -2, -4 \rangle \\ &=& \langle 2-2t, 2-4t,-4t\rangle \end{array}

Moreover, notice that $\ell\left(-1/2\right) = P_3$.

Answer 2

Since it's a line, then you only need one parameter (since it's one-dimensional). I'd go for $X=P_1+t\vec{P_1P_3}$, instead. Your answer isn't incorrect, per se, but once you introduce the extraneous parameter, the points on the line are no longer uniquely described. For example, so long as $t=2s$, the point described is simply $P_1$. We don't want infinitely many ways to describe the same point. One consistent way will be fine.