How are situated the lines $k$ and $m$ in the space $E3$?

Question

Given:

k: $\left\{\!\begin{aligned} & x=1+2t \\ & y=-7+t \\ & z=3+4t \end{aligned}\right.$

m: $\frac{ x+6 }{ 3 } =\frac{ y+1 }{ -2 } =\frac{ z+2 }{ -1 }$

My decision:

Let

m: $\left\{\!\begin{aligned} & x=-6+3p \\ & y=-1-2p \\ & z=-2-p \end{aligned}\right.$

Then:

$\left\{\!\begin{aligned}& 1+2t=-6+3p \\ & -7+t=-1-2p \\ & 3+4t=-2-p \end{aligned}\right. \Leftrightarrow \left\{\!\begin{aligned}& 1+2t+6-3p=0 \\ & -7+t+1+2p=0 \\ & 3+4t+2+p=0 \end{aligned}\right.$

If I solved:

$\left\{\!\begin{aligned} & 2t - 3p = -7 \\ & t + 2p = 6 \end{aligned}\right.$

And find:

$\left\{\!\begin{aligned} & t = \frac{ 4 }{ 7 } \\ & p = \frac{ 19 }{ 7 } \end{aligned}\right.$

but it is not suitable for: $4t+p=-5$. So the system of equations has no solutions.

And I get that: $k \cap m = \varnothing$.

But, do I need to check: $k \parallel m$ and $k \perp m$?

Answer 1

$k:$ $(x,y,z)=(1,-7,3)+t(2,1,4)$ and $m:$ $(x,y,z)=(-6,-1,-2)+s(3,-2,-1).$

Since $$\vec{(2,1,4)}\vec{(3,-2,-1)}=2\cdot3+1\cdot(-2)+4\cdot(-1)=0,$$ we see that $k\perp m$.

Let $k||m$.

Thus, there is $\lambda$, for which $$\vec{(2,1,4)}=\lambda\vec{(3,-2,-1)}$$ or $$2=3\lambda,$$ $$1=-2\lambda$$ and $$4=-\lambda,$$ which is impossible.

Id est, $k$ is not parallel to $m$