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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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    @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!2012-09-12

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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! :)2012-09-13