Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Find the line passing thought the point $p=(1,2,0)$, paralel to the plane $P=\{x,y,z \mid x+2y-z=-4\}$ and crossing the line $L=\{(x,y,z):x+2y=2, y+z=4\}$

So I've tried to put the equation of plane and insert the point, because I think that if line is paralel to the plane, then line is contained in the plane.

But I've checked, that the point don't lay in the plane, because

$x+2y-z=1+4-0=5 \neq -4$

So I'm stuck here.

-- EDIT --

Is that correct:

$x+2y-z=k\,\,\, k\in\mathbb{R}$

Intercepting point $ \begin{cases} x+2y=z\\ x+2y=2\\ x+z=4 \end{cases}\implies\begin{cases} z=2\\ x=2\\ y=0 \end{cases}$

General line equation

$p_{1}-\lambda p_{2}=0 $


$ x=1-2\lambda$



So the answer is



share|cite|improve this question
You find it.${}{}{}{}$ – Alexander Gruber Mar 22 '13 at 9:52
Please have a thorough look at the FAQ about homework questions. – A.P. Mar 22 '13 at 10:02
OK, I'll look now – Joggi Mar 22 '13 at 10:04
@AlexanderGruber Fixed. I don't know how to find it – Joggi Mar 22 '13 at 10:07

It says that the line you need to find is parallel to the given plane, not that it's on that plane.

As this is a homework problem, I'm not going to do it for you, but I will say that you've already done one step, although you may not have realised it.

There are three steps - find plane, find intercept point, construct line through points. I'll leave it to you to figure out what I'm saying from there.

share|cite|improve this answer
How to find intercept points? I need to solve equation system with the $x+y2-z=k$ and equations for line $L$ ? – Joggi Mar 22 '13 at 10:33
This point will be (2,0,2) ? – Joggi Mar 22 '13 at 10:36
You need to find the plane first. You haven't done that, yet. – Glen O Mar 22 '13 at 11:03
If at any point you think you've solved it, check if it satisfies the three conditions - does it pass through the appropriate point? Does it have an intercept with the given line? Is it parallel to the given plane? – Glen O Mar 22 '13 at 11:09

Hint: The possible directions of your line are given by the equation $$ x+2y-z=0 $$

Your solution isn't correct: it is true that $(1,2,0)$ is on your line, but the intersection with $L$ is empty: $$ \begin{cases} 2+z=4\\ (1-z)+4=2 \end{cases} $$ gives $z=2$ (first equation) and $z=3$ (second equation.

Here is one possible solution, using Cartesian equations, like yours (my hint would have provided a solution in parametric equations).

What we're going to do is find the required line as the intersection of two planes: one parallel to the given one and the other containing $L$, both passing through $P=(1,2,0)$.

For the first plane we just have to put $(1,2,0)$ in the equation $x+2y-z=k$, giving $$ \pi_1: x+2y-z=5 $$ Now, the pencil of planes containing $L$ is simply $$ \mathcal{F}: \lambda(x+2y-2)+\mu(y+z-4)=0 $$ thus to find the plane through $P\notin L$ we just need to substitute $(1,2,0)$ $$ \begin{gather} \lambda(1+4-2)+\mu(2-4)=3\lambda-2\mu=0\\ 3\lambda=2\mu \end{gather} $$ therefore our second plane is given by $$ \pi_2:2(x+2y-2)+3(y+z-4)=2x+7y+3z-16=0 $$ hence the required line is given by $$ R:\left\{(x,y,z):x+2y-z=5, 2x+7y+3z=16\right\} $$ You can now check that $R\cap L\neq\varnothing$ and that $P\in R$.

share|cite|improve this answer
So every paralel plane is described by $x+2y-z=0$? – Joggi Mar 22 '13 at 10:23
No, every parallel plane is described by $x+2y-z=k$ with $k\in \Bbb R$ (same orthogonal direction, but passing through different points). If you are taking values in $\Bbb R$, of course. – A.P. Mar 22 '13 at 10:24
Thanks. Could You check if I've done it correctly, in question edit? – Joggi Mar 22 '13 at 10:43
I'm sorry but no, it isn't. – A.P. Mar 22 '13 at 10:56
I provided a full solution in my edit. – A.P. Mar 22 '13 at 11:20

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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