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#1

$Old$ May 20th, 2009, 03:17 AM

qkfxjs $qkfxjs is offline$

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$Default$ distance between two skew lines.

Find the distance between two skew lines:
L1: x = 2s, y = 2s - 1, z = s + 1, and
L2: x = t, y = 2t - 2, z = 3.

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#2

$Old$ May 20th, 2009, 09:40 AM

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Originally Posted by qkfxjs $View Post$

Find the distance between two skew lines:
L1: x = 2s, y = 2s - 1, z = s + 1, and L2: x = t, y = 2t - 2, z = 3.

If $P+tD~\&~Q+sE$ are two skew lines, the the distance between them is $\frac{{\left| {\overrightarrow {PQ} \cdot \left( {D \times E} \right)} \right|}}{{\left\| {\left( {D \times E} \right)} \right\|}}$

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$Old$ May 20th, 2009, 10:53 AM

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$Default$ Ok

So how can I solve the problem.

I'm still having a problem.

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$Old$ May 20th, 2009, 11:15 AM

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Originally Posted by qkfxjs $View Post$

So how can I solve the problem.

Find the direction vector of each line. Those are $D~\&~E$ .

Find a point on each line. Those are $P~\&~Q$

Then use the formula.

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