Annoying vector question.

Try and use similtaneous eqns for the two collinear eqns? And the apply the r=b(lamda)+(1-lamda)a formula?

doesnt have it

how would that work?.

Also the two non-collinear lines cannot intersect.

..

l1 is r = 5i + j + 5k + s(i - j - 2k)
l2 is r = i + 11j + 2k + t(-4i - 14j + 2k )

part 1 ) find shortest distance - done

part 2 ) find vector equation of the line that contains the shortest distance found in part 1 - cant do.

did you get 90/2sqr(259) for shorted distance?

Also wats the answer to part 2?

l1 is r = 5i + j + 5k + s(i - j - 2k)
l2 is r = i + 11j + 2k + t(-4i - 14j + 2k )

part 1 ) find shortest distance - done

part 2 ) find vector equation of the line that contains the shortest distance found in part 1 - cant do.

sorry - the answer isnt in the book.

No I didnt get that distance. Can you explain to me how you got that - i think im wrong.

The shortest distance between two skew lines can be calculated by this:

$\dfrac{(\mathbf{a} - \mathbf{c}) \cdot (\mathbf{b} \times \mathbf{d})}{|\mathbf{b} \times \mathbf{d}|}$, where a and c are the position vectors of the 2 lines and b and d are their direction vectors.

http://www.netcomuk.co.uk/~jenolive/skew.html this may help you!

thats not the problem. The problem is the second part - actually finding the vector equation.

If your two lines are a+sb and c+td.

A direction vector of the line of shortest length is bxd

So, one way is, by going along the first line, and then across the shortest connecting line, you must get to the second line, hence:

a+sb + u(bxd) = c + td where u is a constant.

Solve for s and t, and hence get the two points on the shortest line, and derive its equation from them.

I expect there is a slicker method, but this is doing it from scratch.

If your two lines are a+sb and c+td.

A direction vector of the line of shortest length is bxd

So, one way is, by going along the first line, and then across the shortest connecting line, you must get to the second line, hence:

a+sb + u(bxd) = c + td where u is a constant.

Solve for s and t, and hence get the two points on the shortest line, and derive its equation from them.

I expect there is a slicker method, but this is doing it from scratch.

Sorry do you mind explaining a bit more? How would I get s and t from there? Would we have to find u?