# Vectors - shortest distance between two 3d lines

• Oct 29th 2013, 02:05 PM
jimi
Vectors - shortest distance between two 3d lines
'Show that the shortest distance between the line with equations

(x + 4)/3 = (y - 3)/2 = (z + 6)/5

and the line with equations x - 2y - z = 0, x - 10y - 3z = -7
is 1/2 sqrt 14.'

I rewrote the equations in vector form:

-4i + 3j - 6k + s(3i + 2j + 5k)
7i/2 + 7k/2 + t(-2i + J - 4k)

I found the cross product to be: -13i + 2j + 7k
(-15i/2 + 3j - 19k/2 ).(-13i + 2j + 7k) = 97.5 + 6 - 66.5 = 37
|-13i + 2j + 7k| = sqrt 222
so my answer is 37/sqrt 222.
Can someone please tell me where I'm going wrong?
Thanks.
• Oct 29th 2013, 02:23 PM
Plato
Re: Vectors - shortest distance between two 3d lines
Quote:

Originally Posted by jimi
'Show that the shortest distance between the line with equations

(x + 4)/3 = (y - 3)/2 = (z + 6)/5

and the line with equations x - 2y - z = 0, x - 10y - 3z = -7
is 1/2 sqrt 14.'

If $\displaystyle \ell _1: P+t\overrightarrow D ~\&~\ell _2: Q+t\overrightarrow E$ are two skew lines
then the distance between them is $\displaystyle d({\ell _1},{\ell _2}) = \frac{{\left| {\overrightarrow {PQ} \cdot (\overrightarrow D \times \overrightarrow E )} \right|}}{{\left\| {\overrightarrow D \times \overrightarrow E } \right\|}}$
• Oct 30th 2013, 01:27 AM
jimi
Re: Vectors - shortest distance between two 3d lines
Plato

Yes, that's the formula I'm trying to use, but I don't agree with their answer. Can you tell me where I'm going wrong?
Thanks.