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?

  • $\begingroup$ 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? $\endgroup$ – Monolite Oct 17 '15 at 20:58
  • $\begingroup$ @RoryDaulton sorry I had a mistake in my system, does the question make more sense now? $\endgroup$ – Monolite Oct 18 '15 at 1:04
  • $\begingroup$ Yes, your question makes more sense now. It is also easier to answer: see my answer below. $\endgroup$ – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.