AS Further Maths question Watch

jackwhurt
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Hi, question is how do you find the shortest distance between two skew lines.
Basically, I think I have done it, but I don't know why it works. I found a vector perpendicular to both lines, then found the unit vector of this. Then did the dot product of the unit vector and the difference between points on both lines.
Thanks for any explanation.
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ghostwalker
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(Original post by jackwhurt)
Hi, question is how do you find the shortest distance between two skew lines.
Basically, I think I have done it, but I don't know why it works. I found a vector perpendicular to both lines, then found the unit vector of this. Then did the dot product of the unit vector and the difference between points on both lines.
Thanks for any explanation.
Yes, that's one of the standard methods - assuming I interpret you correctly.

The line of shortest distance between two skew lines will be perpendicular to each of those lines; hence the unit vector your found.

The next bit is a little tricky to see, IMO.

Suppose we have a point p on l1, and a point q on l2.

Then the vector q-p represents the position of the point q on l2 relative to a point p on l1.

Taking the dot product of that with your unit vector is simply finding the magnitude of that vector in the direction of the unit vector, i.e. the perpendicular distance.

Doesn't matter which point you take on l1 or l2.

Hope that helped; but I can see myself that it's not the clearest explanation.
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mqb2766
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There are a few youtube videos about this, have you googled them?

The first part is why must the line (distance) be perpendicular to the two lines? This is basically a double version of the problem of finding the closest distance of a line from the origin. Distance is represented as a circle centred on the origin. As this expands, it will just touch the line, so the forms a tangent to the circle. At this point, the line is perpendicular to the line connecting the origin to the tangent point. Yours is like this but works in both ways for the two skew lines.

See ghostwalkers post after this?


(Original post by jackwhurt)
Hi, question is how do you find the shortest distance between two skew lines.
Basically, I think I have done it, but I don't know why it works. I found a vector perpendicular to both lines, then found the unit vector of this. Then did the dot product of the unit vector and the difference between points on both lines.
Thanks for any explanation.
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RDKGames
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(Original post by jackwhurt)
Hi, question is how do you find the shortest distance between two skew lines.
Basically, I think I have done it, but I don't know why it works. I found a vector perpendicular to both lines, then found the unit vector of this. Then did the dot product of the unit vector and the difference between points on both lines.
Thanks for any explanation.
Yes that's the jist of the formula.

Consider two lines that skew \ell_1 : \mathbf{r} = \mathbf{a} + \lambda \mathbf{c} and \ell_2 : \mathbf{r} = \mathbf{b} + \mu \mathbf{d}.

Now, the shortest distance between these two lines is along a perpendicular vector to both of these. The perpendicular vector is

\mathbf{n} = \mathbf{c} \times \mathbf{d}.


Now consider the vector \mathbf{b}-\mathbf{a} (green vector in the first diagram). This is a vector from \ell_1 to \ell_2, connecting the points on both lines as seen in their equations.

This vector is not necessarily the shortest distance, but we can project it onto the shortest distance by dotting it with the unit normal vector.

Why does this work?

Look at the second diagram. Note that I have shifted the green vector to the right a bit, so that it comes off the same point where the normal vector meets \ell_1. By basic trigonometry,

d = |\mathbf{b}-\mathbf{a}| \cos \theta


But notice, that from the definition of the dot product, we have

(\mathbf{b}-\mathbf{a}) \cdot \mathbf{n} = |\mathbf{b}-\mathbf{a}|  |\mathbf{n}| \cos \theta.... which means that, after dividing both sides by |\mathbf{n}|, we have

|\mathbf{b}-\mathbf{a}| \cos \theta =\dfrac{(\mathbf{b}-\mathbf{a})\cdot \mathbf{n}}{|\mathbf{n}|}


Hence, we have that d = \dfrac{(\mathbf{b}-\mathbf{a})\cdot \mathbf{n}}{|\mathbf{n}|} = (\mathbf{b}-\mathbf{a})\cdot \hat{\mathbf{n}}


where \hat{\mathbf{n}} = \dfrac{\mathbf{n}}{|\mathbf{n}|} is the unit normal vector.




Last edited by RDKGames; 4 weeks ago
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RogerOxon
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(Original post by jackwhurt)
Hi, question is how do you find the shortest distance between two skew lines.
Basically, I think I have done it, but I don't know why it works. I found a vector perpendicular to both lines, then found the unit vector of this. Then did the dot product of the unit vector and the difference between points on both lines.
Thanks for any explanation.
What this boils down to is the distance between two planes with the same normal that contain the directions of both lines.
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ghostwalker
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(Original post by RDKGames)
Yes that's the jist of the formula.

Consider two lines that skew \ell_1 : \mathbf{r} = \mathbf{a} + \lambda \mathbf{b} and \ell_2 : \mathbf{r} = \mathbf{c} + \mu \mathbf{d}.

Now, the shortest distance between these two lines is along a perpendicular vector to both of these. The perpendicular vector is

\mathbf{n} = \mathbf{b} \times \mathbf{d}.
I think you mean:


Consider two lines that skew \ell_1 : \mathbf{r} = \mathbf{a} + \lambda \mathbf{c} and \ell_2 : \mathbf{r} = \mathbf{b} + \mu \mathbf{d}.

Now, the shortest distance between these two lines is along a perpendicular vector to both of these. The perpendicular vector is

\mathbf{n} = \mathbf{c} \times \mathbf{d}.


And carry on as before.
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RDKGames
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(Original post by ghostwalker)
I think you mean:


Consider two lines that skew \ell_1 : \mathbf{r} = \mathbf{a} + \lambda \mathbf{c} and \ell_2 : \mathbf{r} = \mathbf{b} + \mu \mathbf{d}.

Now, the shortest distance between these two lines is along a perpendicular vector to both of these. The perpendicular vector is

\mathbf{n} = \mathbf{c} \times \mathbf{d}.


And carry on as before.
Ah yes, got distracted halfway through making that post and forgot what letters were where
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jackwhurt
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Thanks everyone, I get it now - hopefully will ace my test coming up
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