I have the following two parametric equations of lines:
$$\begin{cases} x = -t + 1 \\ y = t + 3 \\ z = -6t \end{cases} \quad \land \quad \begin{cases} x = 2s + 4 \\ y = -s \\ z = 2s + 1 \end{cases}$$
I want to examinate their mutual position, that is, I want to find out if they intersect.
$$\begin{cases} -t + 1 = 2s + 4 \\ t + 3 = -s \\ -6t = 2s + 1 \end{cases}$$
$$\begin{cases} -t + 1 = 2s + 4 \\ -t – 3 = s \\ -6t = 2(-t-3) + 1 \Rightarrow -6t = -2t – 5 \Longrightarrow t = \frac{5}{4}\end{cases}$$
$$-\frac{5}{4} – 3 = s \Longrightarrow s = – \frac{17}{4}$$
So they intersect for $t = \frac{5}{4} \lor s=-\frac{17}{4}$.
Plugging $t$ into first equation:
$$\begin{cases} x = -\frac{5}{4} + 1 \\ y = \frac{5}{4} + 3 \\ z = -6 * \frac{5}{4} \end{cases}$$
Point $P(-\frac{1}{4}, \frac{17}{4}, -\frac{15}{2})$ is the intersection point.
Thing is, I don't think that's true. I sketched those lines in parametric plotter and they look like this:
They look like skew to me. What went wrong?
Best Answer
Adding the first two equations, $4=s+4$ and obviously $s=0$, then $t=-3$. As this doesn't fit with the third, there is no intersection.
You didn't check that your $s,t$ values satisfy all three equations.