There is enough information. Unless two lines are parallel, there is a unique line of minimum length that joins them.
If $\mathbf{U}$ and $\mathbf{V}$ denote the direction vectors of the two given lines, then you are correct that the minimum-length line is in the direction of the cross product $\mathbf{W} = \mathbf{U} \times \mathbf{V}$.
The first line is defined by the point $\mathbf{P} = (1,2,-1)$ and the vector $\mathbf{U}=(1,-1,1)$, so its equation can be written $\mathbf{A}(s) = \mathbf{P} + s\mathbf{U}$. Similarly, the second line has equation $\mathbf{B}(t) = \mathbf{Q}+t\mathbf{V}$, where $\mathbf{Q} = (4,1,0)$ and $\mathbf{V}=(-2,0,-1)$.
To achieve minumum distance, the "joining" line $\mathbf{A}(s) - \mathbf{B}(t)$ must be perpendicular to the given two lines, so:
$$
\left[(\mathbf{A}(s) - \mathbf{B}(t) \right] \cdot \mathbf{U} = 0
$$
$$
\left[(\mathbf{A}(s) - \mathbf{B}(t) \right] \cdot \mathbf{V} = 0
$$
Substituting for $\mathbf{A}(s)$ and $\mathbf{B}(t)$ and rearranging, we get
$$
s(\mathbf{U} \cdot \mathbf{U}) - t(\mathbf{U} \cdot \mathbf{V}) =
\mathbf{U} \cdot (\mathbf{Q} - \mathbf{P})
$$
$$
s(\mathbf{U} \cdot \mathbf{V}) - t(\mathbf{V} \cdot \mathbf{V}) =
\mathbf{V} \cdot (\mathbf{Q} - \mathbf{P})
$$
These equations have a unique solution unless the original two lines are parallel. If $s_0$ and $t_0$ are the solutions, then the closest points of the two lines are then $\mathbf{A}(s_0)$ and $\mathbf{B}(t_0)$.
the solution is:
$$\vec{r}=\vec{v}t+\vec{OP}$$
in your case is:
$$(x,y,z)=(1,-3,2)t+(-1,1,3)$$
god bless us
Best Answer
Things you need to know to do this problem:
PARAMETRIC FORM
The parametric form of a line is given by $\mathbf r(t) = \mathbf r_0 + t\mathbf v$. Let's forget about the $\mathbf r_0$ for a second: $\mathbf r(t) = t\mathbf v$ is the set of all the points on the line that passes through the origin and the point $\mathbf v$. So we can see that $\mathbf v$ defines the direction of our line -- thus it is called a "direction vector" for the line (but remember, there are an infinite number of direction vectors for any line -- for instance, once you find a vector $\mathbf v$, then $2\mathbf v$ is ALSO a direction vector).
This is a plot of the line that passes through the origin and the point $(1,2)$, i.e. $(x,y) = t(1,2)$:
So then what does $\mathbf r_0$ do? It translates that line. Say you want the line which is parallel to the one given above, but which passes through the point $(-1,1)$, then the equation for it is just $(x,y) = (-1,1) + t(1,2)$. Here's the plot:
So to find the parametric form of the equation of a line you just need one vector (point) which represents the direction of the line -- I used $(1,2)$ in my example -- and one point on the line -- I used $(-1,1)$.
NOTE Your textbook may specify the parametric form as a set of scalar equations. That is exactly equivalent to as my vector equation above, just in a less compact form. In scalar form this second line would be $x=t-1$, $y=2t+1$. Can you see how this is exactly the same?
IMPLICIT FORM
Now, let's consider what you're calling the implicit form (I've never heard it called that, but I am very familiar with it nonetheless). An affine space (like lines and planes and such) given by $\mathbf n \cdot (\mathbf r-\mathbf r_0)=0$ will always be an $(n-1)$-dimensional space, where $n$ is the dimension of the ambient space. For instance, in your problem, you have a line specified in the Euclidean plane. A plane is $2$-dimensional and a line is $1$-dimensional, so we can represent your line via this formula.
Now, what is this formula actually saying? Well, first we need to remember the definition of orthogonality (perpendicularity). If $\mathbf a, \mathbf b \in \Bbb R^n$ (that is $\mathbf a$ and $\mathbf b$ are vectors (points)), then $\mathbf a\ \bot\ \mathbf b$ (this is the symbol for $\mathbf a$ is orthogonal to $\mathbf b$) if and only if $\mathbf a \cdot \mathbf b =0$. So this implicit formula clearly has to do with orthogonal vectors.
Let's consider the line passing through points $\vec a$ and $\vec b$ (I'm changing my notation so that it corresponds to this picture I found). Then clearly $\vec a - \vec b$ is a vector pointing in the direction of our line. See the picture here:
Now, what if instead of the vector $\vec a$, we just put $(x,y)$ -- which is some arbitrary point on our line. Then for $\vec b$ we could put in some specific point $(x_0, y_0)$ (where those would be filled in with numbers) on the line. Then if we had any vector orthogonal to our line, it would have to be orthogonal to $(x,y) - (x_0,y_0)$. So if we had such an orthogonal vector, it would completely specify the conditions that all $(x,y)$ must have to be on the line. Yay.
So let's see how to do it in the case of the line I gave you above (the one going through point $(-1,1)$ with direction vector $(1,2)$). Verify for yourself that an orthogonal vector to the direction vector $(1,2)$, is $(-2,1)$. Then we can just plug in: $\mathbf n \cdot (\mathbf r-\mathbf r_0)=(-2,1)\cdot((x,y)-(-1,1))=0$.
Now we've found the implicit form AND parametric forms for this line. Now can you find them for YOUR line?