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In each case find the shortest distance from the point P to the line, and find the po

  1. Jun 4, 2012 #1
    1. The problem statement, all variables and given/known data
    In each case find the shortest distance from the point P to the line, and find the point Q on the line closest to P.

    P(-1,0,1); [x y z]^T = [3 1 -4]^T +t[3 -2 0]


    2. Relevant equations

    Projection equation

    3. The attempt at a solution
    Let P_o be (3,1-4)

    Vector V = Vector P_o to P = (-1,0,1) - (3,-1,4) = (-4,-1,-3)

    Vector V_1 = projection of vector V over direction vector
    = ( ( (-4,-1,-3)·(3,-2,0) ) / (sqrt. ((3)^2 + (-2)^2)) (3,-2,0)
    = (-10 / 13) (3,-2,0)

    ||Vector QP|| = || vector V - vector V_1 ||
    = || (-4,-1,-3) + (30/13,-20/13,0)||
    =|| (-52/13,-13/13,-3) + (30/13,-20/13,0)||

    = ||(-22/13,-33/13,-3/13)||
    = (1/13) (sqrt.1582)

    In the back of the book it says (1/13) (sqrt.1846). Where did I go wrong?
     
    Last edited: Jun 4, 2012
  2. jcsd
  3. Jun 4, 2012 #2

    HallsofIvy

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    Science Advisor

    Re: In each case find the shortest distance from the point P to the line, and find th

    How did "(3, 1, -4)" before become (3, -1, 4)? Was that a typo? Unfortunately, that error propogates. The vector from P_0 to P is (-1, 0, 1)- (3, 1, -4)= (-4, -1, 5), not (-4, -1, -3).

    Personally, I would have done this in a completely different way: the plane through (-1, 0, 1) perpendicular to the given line is 3(x+ 1)- 2(y- 0)+ 0(z- 1)= 3x+ 3- 2y= 0 or 3x- 2y= -3. The line, x= 3+ 3t, y= 1- 2t, z= 4, crosses that plane when 3(3+ 3t)- 2(1-2t)= 9+ 9t- 2+ 4t= 13t+ 7= -3 so that t -10/13. That is, x= 3+ 3(-10/13= (39- 30)/13= 9/13. y= 1- 2(9/13)= (13- 18)/13= -5/13, z= 4. Find the distance from (-1, 0, 1) to (9/13, -5/13, 4).
     
  4. Jun 4, 2012 #3
    Re: In each case find the shortest distance from the point P to the line, and find th

    Yes, it was a typo. The point should be (3, -1, 4) , and not (3, 1, -4).

    For the the line, x= 3+ 3t, y= 1- 2t, z= 4 , should 1 or 4, in y and z respectively, be negative? (since the point, at the time you typed your response, could have been (3, -1, 4) or (3, 1, -4).

    Thank you for your help. I will try your method and see if I get the correct answer.
     
  5. Jun 4, 2012 #4
    Re: In each case find the shortest distance from the point P to the line, and find th

    Using your method I almost obtained the correct answer.

    P(-1,0,1); [x y z]^T = [3 -1 4]^T +t[3 -2 0]

    3(x-1) - 2(y-0) - 0(z-1)
    = 3x -3-2y
    or
    -3 = 3x -2

    and
    x = 3+3t
    y = -1-2t
    z = 4

    ---

    -3 = 3(3 +3t) - 2(-1-2t)
    -3 = 11 + 13t
    -14/13 = t

    ---

    x = 3 - 42/13 = -3/13
    y = -1 + 28/13 = 15/13
    z = 4

    (-3/13, 15/13, 4) <------- Point Q
    ---

    (-1,0,1) - (-3/13, 15/13, 4)
    = (10/13, 15, 3)

    ||(10/13, 15, 3)||
    =(1/13)(sqrt1846) <------- Distance from Point Q to Point P

    --
    So I obtained the correct magnitude for QP and point for Q just like in the back of the book.

    Thank you very much for your assistance!
     
    Last edited: Jun 4, 2012
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