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
Best Answer
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.