My Math Forum shortest distance between 2 lines in 3-space

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 October 31st, 2013, 01:08 AM #1 Newbie   Joined: Oct 2013 Posts: 17 Thanks: 0 shortest distance between 2 lines in 3-space 'Show that the shortest distance between the line with equations: $\frac{x+4}{3}=\frac{y-3}{2}=\frac{z+6}{5}$ and the line with equations: $x-2y-z=0$ $x-10y-3z=-7$ is $\frac{1}{2}\sqrt14$' I'm trying to use the formula $\frac{(a-c).(bxd)}{|bxd|}$ I take a to be -4i + 3j - 6k (position vector of a point on line 1) c to be 7i/2 + 7k/2 (position vector of a point on line 2) b is 3i + 2j + 5k (direction of line 1) d is -2i + j - 4k (direction of line 2) I make a - c to be -15i/2 + 3j - 19k/2 and b x d to be -13i + 2j + 7k I make (a - c).(b x d) = 37 and |b x d| = $\sqrt{222}$ so my answer is: $\frac{37}{\sqrt{222}}$ which does not agree with the question. Can someone please tell me where I am going wrong? Thank you.
 October 31st, 2013, 04:25 AM #2 Senior Member   Joined: Jun 2013 From: London, England Posts: 1,312 Thanks: 115 Re: shortest distance between 2 lines in 3-space I agree with you. Unless you've mistyped something in the original equations.
 November 4th, 2013, 12:21 AM #3 Newbie   Joined: Oct 2013 Posts: 17 Thanks: 0 Re: shortest distance between 2 lines in 3-space Thank you Pero, much appreciated.

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