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Homework Help: 4D: mimimum distance between point and parametric line

  1. Aug 13, 2010 #1
    1. The problem statement, all variables and given/known data
    There is points:
    P=(1,2,3,4)
    Q=(4,3,2,1)
    and line L passes through P and is parallel to A
    A=(1,1,1,1).
    X(t) is anypoint on line L.

    1. Find the distance between X and Q as a fucntion of t.
    2. Find the minimum distance between Q and the line.(which is 2(51/2))


    2. Relevant equations
    Parametric equations....
    distance between two points.

    3. The attempt at a solution

    Im really not sure if my approach is right at all.... i just want to understand where im going wrong.

    For 1.
    I tried the parametric equation of the line which is
    X(t)=P+At

    to go from X(t) to Q its
    X(t)-Q=(P-Q)+At=(-3,-1,1,3)+(1,1,1,1)t
    X(t)-Q=(-3+t,-1+t,1+t,3+t)
    the distance between X(t) and Q would be
    d=||(X(t)-Q)|| or d=((X(t)-Q).(X(t)-Q))1/2


    For 2.
    then the distance squared as a function of t would be
    d2=(X(t)-Q).(X(t)-Q)

    this gives d2=20+4t2
    the derivative is (d2)'=2((d2)'d(2
    (d2)'=(2(8t)(20+4t2)
    (d2)'=320t+64t3

    set the derivative to 0 for minimum and i get
    t=(-320/64)1/2

    Its clearly not the right answer of 2(5)1/2

    I dont know where im going wrong.
     
    Last edited: Aug 13, 2010
  2. jcsd
  3. Aug 13, 2010 #2

    hunt_mat

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    You can also minimise d^2 to get the shortest distance between a point and a line. This is just the normal from the line, can you just use that?

    Mat
     
  4. Aug 13, 2010 #3
    I've minimised d^2 by taking the derivative and setting to 0
     
  5. Aug 13, 2010 #4

    vela

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    You differentiated incorrectly. What Mat is suggesting is that you minimize f(t)=20+4t2. There's no need to use the chain rule.
     
  6. Aug 13, 2010 #5
    How can you minimize f(t)=20+4t2?
    Is it not only do when is a minus?
     
  7. Aug 13, 2010 #6

    vela

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    You can do it by inspection. It's a parabola, right?

    Or you can solve f'(t)=0. Note I just called it f(t) instead of d2 since the notation seemed to be leading you astray.
     
  8. Aug 13, 2010 #7

    HallsofIvy

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    y= 20+ 4t2 is a parabola opening upward. It's minimum value is at the vertex (which is very easy to find here- t2 is never negative).

    You may be thinking that y= 20+ 4t2 has no maximum- and that y= 20- 4t2 has no minimum.
     
  9. Aug 13, 2010 #8
    so t=5 is the time where the distance between the line is at a minimum?
    .....visualisation helps sometimes

    the substitue t into X(t)

    and then find distance betweeb X(t) and Q?
     
  10. Aug 13, 2010 #9
    sorry i mean t=20
     
  11. Aug 13, 2010 #10

    vela

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    No. How did you get t=20?
     
  12. Aug 13, 2010 #11
    I see now i get what u mean by minimize.
    Find the value of d^2 which is the smallest.

    The smallest value of d^2 is 20.
    so d=2(5)^1/2
     
    Last edited: Aug 13, 2010
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