I've been given this problem but I'm not sure there is enough information to solve it, I was wondering if anyone could verify that. Thank you.
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Let r be a line going through the points $(1,2,-1)$ and $(2,1,0)$ and let s be a line going through the points $(4,1,0)$ and $(2,1,-1)$.
Find the equation of a line t, perpendicular to both lines and common to both (it "touches" both of them).
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You could find the vector of line t by cross product from the lines r and s, but how would you find the exact equation of the line? It doesn't say that the line t goes through neither of the points mentioned.
Thank you,
Bruno
Best Answer
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 $\mathbf{B}(t_0)$.