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Problem

Given two lines: $$(l_1)=(x,y,z)=(3,-1,2)+t(1,1,0)$$ $$(l_2)=(x,y,z)=(0,5,2)+t(1,-2,1)$$ Explain why these two lines are not parallel and why they do not intersect each other. Also, find the minimum distance between these two lines.

Progress

I'm pretty sure you have to find the projection of a vector on one line onto the a vector joining the two but I'm not sure how to go about it. Direction vector of the first line is given: $(1,1,0)$. Where will the joining vector come from?

Any help would be great.

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  • $\begingroup$ It's usually polite to actually ask a question when asking a question, rather than giving a command. Additionally, you should really show that you've at least tried to work out how to get the solution before posting. $\endgroup$ – ahruss Feb 14 '13 at 2:15
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For two lines to be parallel, the vectors defining their slopes have to be parallel, which means their vectors have to be constant multiples of one another. It's easy to see by inspection that $\langle1,1,0 \rangle$ is not parallel to $\langle1,-2,1\rangle$. This shows that the lines aren't parallel. To prove they don't intersect, set the two lines' equations equal to one another. This gives $$ \langle3,-1,2\rangle + t\langle1,1,0 \rangle = \langle0,5,2\rangle+s\langle1,-2,1\rangle $$ Solving this component-wise leads to the system $$ \begin{align} 3&=s-t\\ -6&=-2s-t\\ 0&=s \end{align}, $$ which is clearly not possible because substituting $s=0$ as given by the last equation makes the first two equations $t=-3$ and $t=6$, which is a contradiction, so the system has no solution, so the lines don't intersect.

For the third part, the distance, there are two simple ways to solve the problem. First, you can use the distance formula and then minimize it. Let $D(s,t)$, where $s$ is the parameter for $l_2$ and $t$ is the parameter for $l_1$, be the square of the distance between arbitrary points on $l_1$ and $l_2$. Then minimizing $$ D(s,t) = ((3+t)-s)^2 + ((-1+t)-(5-2s))^2 + (2-(2+s))^2$$ gives the minimum distance between the two lines.

Alternatively, and much more easily, you can find a vector $\vec n$ that is perpendicular to both $l_1$ and $l_2$. Then, the distance between $l_1$ and $l_2$ is just the projection of $\overrightarrow{Q_1Q_2}$ onto $\vec n$, where $Q_1$ is a point on $l_1$ and $Q_2$ is a point on $l_2$. You can find $\vec n$ by taking the cross product of $\vec{v_1} = \langle1,1,0 \rangle$ and $\vec{v_2} = \langle1,-2,1\rangle$. That is, $$ \vec n = \vec v_1 \times \vec v_2. $$

Using that value, the minimum distance between the lines is $$ \mbox {proj}_\overrightarrow{Q_1Q_2}\vec n = \frac{\|\overrightarrow{Q_1Q_2}\cdot\vec n\|}{\|\vec n\|} $$

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  • $\begingroup$ perfect thanks! $\endgroup$ – Chance Feb 14 '13 at 2:34
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They are nit parallel because the vector $(1,1,0)$ is indipendent from the vector $(1,-2,1)$. If you write cartestian equations you have $x=3+t,y=t-1,z=2$, so $t=y+1, x=3+(y+1)=y+4,z=2$ for $l_{1}$. For $l_2$ we have $x=t,5-2x,z=2+x$. So we have two systems of equation: $\{x=y+4; z=2\}$ and $\{y+2x=5; z=2+x\}$ and this two systems haven't any common solution. Infact if $z=2$ then $x=0$ and $y=-4$. But if $y+2x=5$ and $x=0$ we have $y=5$. For the distance you have only apply the formula.

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