# Showing that lines from this family do not intersect for different $\alpha$.

I want to show that lines from this family in euclidean 3 dimensional space

$$\begin{cases} \alpha(x-y) + z = 0 \\ x+y+2 \alpha = 0 \end{cases}$$

do not intersect for different values of $\alpha$.

This is a family of lines because it's the intersection of two planes, I think they do not intersect because each line is perpendicular to the $z$ plane at the value of $z = -\alpha(x-y)$ is this enough to conclude?

And this got me thinking how would I represent a family of lines all intersecting in one point in 3 dimensional euclidean space? Could I have a link or a source where to study the general derivation of parallel family of lines in 3 dimensional space?

• Is @NormalHuman a Bot? It had amazing reaction time commenting this question and the name is vaguely suspicious. If not I am sorry but do not understand what tag is wrong/should add? – Monolite Oct 17 '15 at 20:58
• @RoryDaulton sorry I had a mistake in my system, does the question make more sense now? – Monolite Oct 18 '15 at 1:04
• Yes, your question makes more sense now. It is also easier to answer: see my answer below. – Rory Daulton Oct 18 '15 at 11:19

With your last change to your question, we can answer it easily now. Take a point $(r,s,t)$ that satisfies your set of equations for both $\alpha_1$ and $\alpha_2$. By the second equation and considering $\alpha_1$,
$$r+s+2\alpha_1=0$$ so $$\alpha_1=\frac{r+s}{-2}$$
Similarly, considering $\alpha_2$,
$$r+s+2\alpha_2=0$$ so $$\alpha_2=\frac{r+s}{-2}$$
Hence $\alpha_1=\alpha_2$. To summarize, if a point is on two members of your family of sets, those two members are the same set. The contrapositive of this is: Two different members of your family of sets have no element in common. QED.