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Vector Calculus - Having trouble finding a Line

  1. Sep 7, 2008 #1
    I know this has been posted before, and i've read the post concerning the same problem and i've googled this a million times, but i cant seem to get it. So here's the problem:

    Given a Point and a Line, find the Line that passes through the point (3,1,-2) and is perpendicular to the line x=-1+t, y=-2+t, z=-1+t.

    Okay, so i found the direction vector of the given line (which i'll call l) to be (1,1,1).
    I know that a vector lying on l and a vector lying on the line i need to find (which i'll call l'), when you apply the dot product to those 2 vectors it must equal 0 in order for them to be perpendicular.

    I know i have to use orthagonal projection, well i think i need to. So far i took the point given (which i called Q) and a point on the line given (which i called P) to find the vector QP, which (in my case) is (4,3,-1).

    From here i'm lost with what to do, i've been sitting here racking my brain trying different things but i just cant seem to figure out how to find the line. I'm not asking for a solution but a *simple* set of steps or something like that to follow would be greatly appreciated.

    Thanks!
     
  2. jcsd
  3. Sep 7, 2008 #2

    mathman

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    Try the following system:
    given line L=(-1,-2,-1)+(1,1,1)t
    other line K=(3,1,-2)+(a,b,c)s

    Known equations:
    a+b+c=0 (perpendicular)
    a^2+b^2+c^2=1 (normalization, since an arbitrary multiple is allowed)
    To complete the problem definition, we need to assume K and L must meet at some point, otherwise, K could be any line in a plane passing through (3,1,-2) and perpendicular to L. This then gives us 3 more equations K(s)=L(t).

    Solve the system (5 equations - 4 linear and 1 quadratic) for 5 unknowns (a,b,c,s,t).

    Good luck!
     
  4. Sep 7, 2008 #3
    Thanks! I think i get it now and i'll try it in a minute i just have one question. What do you mean by the second step of your "Known Equations" section? I dont see where this one equation is coming from, but i get the rest. Thanks so much!
     
  5. Sep 7, 2008 #4
    Also, i dont know where to begin solving this system of equations :(
     
  6. Sep 8, 2008 #5

    mathman

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    a^2+b^2+c^2=1 is thrown in because (a,b,c) can be replaced by (fa,fb,fc), where f is arbitrary. Otherwise the whole system floats.

    As far as solving the system, I suggest solving the four linear equations to get rid of s and t and either a,b, or c. Since the relationships among a,b, and c will be linear, these can be inserted into the normalization equation to get a quadratic in one remaining variable.
     
  7. Sep 12, 2008 #6
    Thanks so much for the help! I figured it out, our TA did it differently but i prefer the approach suggested - it's more difficult but more concrete. Our TA suggested minimizing the distance equation between the point and the line, which is also effective, i thought of doing that but i liked the more difficult approach suggested. Thanks again!
     
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