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Thread: A problem related with finding the distance in 3-D.

  1. #1
    Sep 2010

    A problem related with finding the distance in 3-D.

    This is my problem:

    ** Find the distance of R(10,-1,4) from the line x=5+4t, y = -2+2t, z=7-4t.
    providing that I have to use the Component to find d.
    And here is the pic of my first steps:

    The thing that I'm stuck is that which vector should I use at the final step (the component)? I tried to use QR and then QS, but surprising enough (for me), for both attempts d becomes 0, which I don't think it should be.

    Then I tried way 2, which is find the equation of the plane containing QR and QS, and I got x + 4y + 3z = 18, which then gives me the same n-vector of <1,4,3>. However, if I choose a point P (say 0,0,18) from that plane, then do the vector PR, I got <-10,-1,14> and then plug that in the component, I got d to be 28/√26. Now the result is a number, that a right way to do?

    so my question is "Which approach is right?" And by the way, would somebody please give me general step(s) of finding the distance (from point to line, point to plane, plane to plane and line to plane)? 'cause my teacher doesn't allow us to remember the formula. She makes us derive it every time by using direction vector v and orthogonal vector n only. But I got mess up btw "when should I cross the vectors" or "when should I use the direction vector instead of normal vector", that kind of questions

    Please help me, I'll really appreciate.
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  2. #2
    MHF Contributor

    Aug 2006
    Frankly, given your teacherís restrictions on formulas, it is difficult to help you.

    I can give you the idea. Given a line \ell :P + tD and that R\notin \ell then the distance from R to \ell is d(R;\ell ) = \left\| {\overrightarrow {PR}  - \frac{{\overrightarrow {PR}  \cdot D}}{{D \cdot D}}D} \right\| = \dfrac{{\left\| {\overrightarrow {PR}  \times D} \right\|}}{{\left\| D \right\|}}.
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