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Math Help - Shortest distance between 2 skew lines

  1. #1
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    Shortest distance between 2 skew lines

    Given 2 skew lines

    ru = (x1, y1, z1) + u(a1, b1, c1)
    rv = (x2, y2, z2) + v(a2, b2, c2)

    Verify the formula for the shortest distance of the line.
    Shortest distance between 2 skew lines-shortest-distance.png

    Okay what I did was that I found the distance between 2 points r = (x1-x2, y1-y2, z1-z2) and then generated a vector that is orthogonal to the 2 lines using cross product and projected r onto d (the distance). The problem I'm having is simplying the equation into the final equation. How do I simplify r.(u x v) into that 3x3 matrix?
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  2. #2
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    Re: Shortest distance between 2 skew lines

    Quote Originally Posted by aqualiary View Post
    Given 2 skew lines
    ru = (x1, y1, z1) + u(a1, b1, c1)
    rv = (x2, y2, z2) + v(a2, b2, c2)
    Verify the formula for the shortest distance of the line.
    Click image for larger version. 

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    I can give you the usual formula.
    Suppose that P+tD~\&~Q+tE are two skew lines, then the distance between them is:
    \frac{|(P-Q)\cdot(D\times E)|}{\|D\times E\|}.

    Now the denominator of that expression is the length of D\times E.
    Do you know how to find D\times E using a determinate?
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  3. #3
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    Re: Shortest distance between 2 skew lines

    So basically I have (P-Q).(D X E) down to 3 2x2 matrices (each adding the other), how do you put that into a 3x3 form?
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    Re: Shortest distance between 2 skew lines

    Quote Originally Posted by aqualiary View Post
    So basically I have (P-Q).(D X E) down to 3 2x2 matrices (each adding the other), how do you put that into a 3x3 form?
    The first row is the entries in P-Q.
    The the next to rows are the usual D\times E.
    That is known as triple product.
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