Annoying vector question. Watch

imzir
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How do you find the vecor equation of the line that runs through the shortest distance between two non-collinear vectors.
I can find out the gradient using cross product but I dont know how to get a coordinate.

Apparently there are more than one method of doing these styles of questions and I cant understand any.

Can someone explain to me how you approach this.

Thank you
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Khodu
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Try and use similtaneous eqns for the two collinear eqns? And the apply the r=b(lamda)+(1-lamda)a formula?
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imzir
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(Original post by Khodu)
Try and use similtaneous eqns for the two collinear eqns? And the apply the r=b(lamda)+(1-lamda)a formula?
how would that work?.

Also the two non-collinear lines cannot intersect.
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boromir9111
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http://en.wikipedia.org/wiki/Plane_%28geometry%29
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imzir
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doesnt have it
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boromir9111
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(Original post by imzir)
doesnt have it
Do you have a question you're working on right now? post it?
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Khodu
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(Original post by imzir)
how would that work?.

Also the two non-collinear lines cannot intersect.
sorry, just realised what I said.

Collinear means when they're both on the same line. So they both have the same direction vector?

http://www.maths.qmul.ac.uk/~sb/MTH4103/Chap06.pdf

6.2 has notes on collinear, hope that helps. I'm currently revising something else br otherwise I'd help. :yy: goodluck
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imzir
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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.

(Original post by boromir9111)
..
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Freakonomics123
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(Original post by imzir)
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?
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imzir
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(Original post by Freakonomics123)
did you get 90/2sqr(259) for shorted distance?

Also wats the answer to part 2?
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.
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boromir9111
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(Original post by imzir)
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.
http://www.netcomuk.co.uk/~jenolive/skew.html this may help you!
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dknt
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(Original post by imzir)
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.
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imzir
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(Original post by dknt)
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.
thats what i used. But what about the second part?
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imzir
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thats not the problem. The problem is the second part - actually finding the vector equation.
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ghostwalker
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(Original post by imzir)
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.
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imzir
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(Original post by ghostwalker)
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.
Thanks that makes a lot of sense too (when I draw a diagram).

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dknt
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(Original post by ghostwalker)
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? :rolleyes:
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ghostwalker
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(Original post by dknt)
Sorry do you mind explaining a bit more? How would I get s and t from there? Would we have to find u? :rolleyes:
By equating coefficients of i, then j, and then k, you end up with three equations in three unknowns.
You're not actually interested in u (since you've already worked out the shortest length), but if you wanted you can use u|bxd| as the shortest length.
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