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.

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 $\displaystyle d({\ell _1},{\ell _2}) = \frac{{\left| {\overrightarrow {PQ} \cdot (\overrightarrow D \times \overrightarrow E )} \right|}}{{\left\| {\overrightarrow D \times \overrightarrow E } \right\|}}$

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.