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$$l_1=(0,1,1)+t(-1,1,0)|t\in\mathbb{R}$$

$$l_2=(4,0,1)+s(-2,2,0)|s\in\mathbb{R}$$

Now, we note that the direction vector for $l_2$, is $2\times (-1,1,0)$, the direction vector of $l_1$.

Hence we can say that the lines are parallel, and so they never intersect.

Is this enough to prove that a plane does not exist that contains both lines? I haven't attempted to find their point of intersection, but would that be nessesary here?

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Two distinct parallel lines always determine a plane.

Clearly, they have the same direction. And since there is no value of $t$ for which $l_1=(4,0,1)$ they are distinct parallel lines.

And the plane in which they lie has no $z$-coordinate other than $z=1$ so the equation of the plane containing both lines is $z=1$

Addendum: Any point in the plane containing the two lines can be expressed in the form $p=(1-c)l_1+cl_2$ for some $c,\,s,\,t$. The $z$ coordinate of $p$ will be $z=[(1-c)\cdot1+(1-c)t\cdot0]+[c\cdot1+cs\cdot0]=1-c+c=1$.

Second Addendum: Since the OP is having difficulty with the short-cut, I will re-do it as if there were no short-cut in this case.

First, we find the nearpoint on $l_1$ to the point $(4,0,1)$. There are several ways to do that, but we can begin by finding the plane through the point $(4,0,1)$ and perpendicular to $l_2$:

\begin{eqnarray} -1(x-4)+1(y-0)+0(z-1)&=&0\\ x-y&=&4 \end{eqnarray}

Next, find the value of $t$ which places the point $(-t,1+t,1)$ of $l_1$ on the plane $x-y=4$.

\begin{eqnarray} x-y&=&4\\ -t-(1+t)&=&4\\ t&=&-\frac{5}{2} \end{eqnarray}

So the nearest point on $l_1$ to the point $P=(4,0,1)$ on $l_2$ is the point $Q=\left(-\frac{5}{2},-\frac{3}{2},1\right)$

So $\vec{PQ}=\left(\frac{13}{2},\frac{3}{2},0\right)$. However, any multiple of this vector will do as well, so we use $\vec{u}=(13,3,0)$ and cross it with the direction vector $v=(-1,1,0)$ to obtain a normal vector $n=(0,0,16)$ for the plane containing both $l_1$ and $l_2$. This gives the equation

\begin{eqnarray} 0(x-4)+0(y-0)+16(z-1)&=&0\\ z&=&1 \end{eqnarray}

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  • $\begingroup$ Can you expand on your last sentence a bit? How do you know about the $z$ coordinate being 1? and the others being $0$? $\endgroup$ – K Split X Oct 22 '17 at 23:57
  • $\begingroup$ OK, hope that helps. $\endgroup$ – John Wayland Bales Oct 23 '17 at 0:02
  • $\begingroup$ I don't understand the part where you say "and the others being 0." $\endgroup$ – John Wayland Bales Oct 23 '17 at 0:11
  • $\begingroup$ Well you say that the equation of the plane is $z=1$, which implies that $z=0(x-x_0)+0(y-y_0)+1$, and so I was wondering why the $0$ came to be? $\endgroup$ – K Split X Oct 23 '17 at 0:18
  • $\begingroup$ For that you would have to work it the long way where you, for example, find the near point $P$ on $l_1$ to $Q=(4,0,1)$ on $l_2$ and find normal vector $n=\vec{PQ}\times(-1,1,0)$ to get $a,b,c$ satisfying $a(x-4)+b(y-0)+c(z-1)=0$. But it is unnessessary to do that in this case since both of the parallel lines have only $1$ for a $z$-coordinate. That in itself guarantees that the equation will simplify to $0(x-4)+0(y-0)+c(z-1)=0$. $\endgroup$ – John Wayland Bales Oct 23 '17 at 0:28
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This is a supplement to John Wayland Bales’ answer.

Working with homogeneous coordinates in extended Euclidean space, a.k.a. the projective space $\mathbb{RP}^3$, the parametric equations of the two lines immediately give us two pairs of points that lie on these lines, namely, $(0:1:1:1)^T$, $(-1:1:0:0)^T$, $(4:0:1:1)^T$ and $(-2:2:0:0)^T$. Now, a plane can be represented by a nonzero homogeneous vector $\mathbf\pi$: all points $\mathbf p$ on the plane satisfy the equation $\mathbf\pi^T\mathbf p=0$. Therefore, if there is a unique common plane, the matrix $$\begin{bmatrix}0&1&1&1\\-1&1&0&0\\4&0&1&1\\-2&2&0&0\end{bmatrix}$$ must have a nullity of $1$. It’s fairly easy to see that this is so by inspection, or one can carry out the usual row-reduction process to obtain $$\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&1&1\\0&0&0&0\end{bmatrix}.$$ From this we can read that the common plane is $\mathbf\pi=(0:0:1:-1)^T$, i.e., the plane $z=1$.

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