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I have two lines:

$r:\begin{cases}x=t \\ y=2t+1\\z=t-2 \end {cases}$

and

$s:\begin{cases}x+2y-z=0 \\ 3x+y+z+1=0 \end {cases}$

I have to establish their relative position.

I have thought that $s=\{v: v \perp u_1\ \text{and } v \perp u_2\}$ ($u_1=(1,2, -1) \text { and } u_2=(3, 1, 1)$). And so I found the vector $w_1$ that spans s: $w_1=(-3, 4, 5)$.

If r//s, their vectors of direction have to be linearly dependents.

The vector of direction of r is $w_2=(1, 2, 1)$.

Well, $w_1 \text { and }$ $w_2$ aren't linearly dependent, so r and s aren't parallel.

But, if I try to solve this set: $\begin{cases}x=t \\ y=2t+1\\z=t-2\\x+2y-z=0 \\ 3x+y+z+1=0 \end{cases}$

I obtain no solutions exist" and so r seems parallel to s.. :/

Where and why am I wronging? :(

Thank you

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Your $\,w_1\,$ does not span $\,s\,$ as it doesn't satisfy the second linear equation defining $\,s\,$ –  DonAntonio Sep 12 '12 at 11:39
    
@DonAntonio I have solved the homogeneous linear system associated by s. I have obtained $x=(-3/4)y$ and $z=(5/4)y$. So assuming $y=b$, I have $x=(-3/4)b$, $y=b$, $z=(5/4)y$ and the space of solutions is spanned by $(-3/4, 1, 5/4)$. Multiplying for 4, I obtain $w_1$. What's wrong? –  sunrise Sep 12 '12 at 11:59
    
just substitute that value into the second equation and it won't satisfy it...! The mistake is that you cannot multiply the solution you obtained as the second equation is not linear: there is that $\,1\,$ there, see? –  DonAntonio Sep 12 '12 at 12:02
    
BTW, neither the vector $\,(-3/4\,,\,1\,,\,5/4)\,$ is a solution... –  DonAntonio Sep 12 '12 at 12:04
    
@DonAntonio Ok, let's restart.. :( You're right.. But I can't understand(!): vectors belonging to s are perpendicular to $u_1$ and $u_2$, isn't it? If I consider a generic v belonging to s, I have that scalar products between v and $u_1$ and between v and $u_2$ have to be 0. And so I have a set of two equations (my two conditions about inner product)... I solve it and.. what's wrong? thanks! –  sunrise Sep 12 '12 at 12:22
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1 Answer

up vote 2 down vote accepted

You correctly obtained the direction $w_1$ of the second line as a vector proportional to $=u_1\times u_2$, and correctly find that the two lines have no intersection.

But you forget that in 3-dimensional space, two lines can be neither parallel nor incident, the third possibility being that they are skew lines.

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All right! Thanks a lot! :) –  sunrise Sep 13 '12 at 19:27
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