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Distance between a point and a line.

  1. Nov 6, 2013 #1
    1. The problem statement, all variables and given/known data

    Find the distance from the point (5, 1, 5) to the line x = 0, y = 1 + 1t, z = 5 + 3t

    2. Relevant equations

    I was thinking of the equation I learned in Linear Algebra but that one only relates a line to a point in 2d, when it falls in 3d it relates a plane to a point...


    3. The attempt at a solution

    I was thinking of using my normal vector ( 0,1,3) to then find a point on my line because I already have one point (which is (0,1,5)) But I get stuck afterwards ... :/
     
  2. jcsd
  3. Nov 6, 2013 #2

    Dick

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    Look at the vector representing the difference between (5,1,5) and a point on your line (0,1+t,5+3t). You want to minimize the length of that vector. What should t be? What kind of course is this?
     
  4. Nov 6, 2013 #3
    Cal 3. It's weird we already did vector function and all, this just happened to be in my online assignment. Anyway, apart from that,

    what do you mean by minimizing? And on the line I have a vector <0,1,3> ( don't know why I said normal).
     
  5. Nov 6, 2013 #4

    Dick

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    If it's calculus then you shouldn't have any problems with minimizing and you don't need any formulas. The distance from (5,1,5) to the line is the distance to the point on the line that minimizes the distance from (5,1,5) to (0,1+t,5+3t). Set it up as a minimization. It seems to be deliberately set up to super easy.
     
  6. Nov 6, 2013 #5
    I have never heard of minimization. I mean we used gradients, partial derivatives, vector derivative, etc... but never have I done minimization, sorry this might sounds somewhat stupid but it's not a term we've used in class
     
  7. Nov 6, 2013 #6

    Dick

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    The distance from a point to a line is the smallest distance between the point and any point on the line. What's the distance between the point and a point on the line? It's a function of t, right? Now find the smallest distance, by taking a derivative, setting it to zero, etc, etc. It's easier if you extremize distance squared rather than distance.
     
    Last edited: Nov 7, 2013
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