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Finding perpendicular distance of a point from line l (using vector equation).

  1. May 14, 2012 #1
    1. The problem statement, all variables and given/known data

    Find a vector equation of the line l containing the points (1,3,1) and (1,-3,-1). Find the perpendicular distance of the point with coordinates (2,-1,1) from l.

    2. The attempt at a solution

    Let (1,3,1) = a, let (2,-1,1) = q. Let N be the point where ANQ = 90 degrees.

    r = (1,3,1) + t(1-1,-3-3,-1-1) = (1,3,1) + t(0,-6,-2)

    |p| = sqrt (6^2 + 2^2) = 2 sqrt 10

    Unit vector, u = (-3 sqrt 10/10)i + (- sqrt 10/10)j

    Vector AQ = q - a = i - 4j

    AN = (Vector AQ).u = 6 sqrt 10/5


    Hence, perpendicular distance NQ = sqrt (|Vector AQ|^2 - AN^2) = 17 - (6 sqrt 10/5)^2 = sqrt 65/5.

    3. Relevant equations

    The book gives the answer AN = 21 sqrt 15/5. Can anyone please check my answer for any mistakes? Thanks.
     
  2. jcsd
  3. May 14, 2012 #2

    HallsofIvy

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    What is "ANQ"?

    Yes, this is the equation of the line l.

    You didn't say, but I guess "p" is the vector from (1, 3, 1) to (1, -3, 1).

    This is a unit vector in the direction of vector -6i- 2j but where did that vector come from? Your line is in the direction of the vector -6j- 2k.

    You keep using notation you haven't defined! What are A and Q? Presumably it is the vector from point a to point q but what is point a?

    I really can't tell what you are doing. What I would do is find the equation of the plane perpendicular to line, l, through (1, 3, 1) and (1, -3, -1), and containing the point (2, -1, 1). Then find the point where line l crosses that plane and find the distance from that point to (2, -1, 1).
     
    Last edited: May 15, 2012
  4. May 15, 2012 #3
    Your numerical answer is good but, as HallsofIvy says, your working needs more clarity. The book answer as given is wrong for the problem as given, assuming I understand your notation (function brackets would be nice for sqrt(15) eg.).

    Can you calculate the coordinates of N, which I take to be the nearest point on the line l to (2,-1,1)?
     
  5. May 18, 2012 #4
    REVISION

    NOTE: Small alphabets in bold indicate vectors. If A is a point, a represents its position vector.

    Let (1,3,1), (1,-3,-1) and (2,-1,1) be the points A, B and Q respectively. Let N be the foot of the perpendicular from Q to the line.

    r = a + tp
    r = (1,3,1) + t(1-1,-3-3,-1-1) = (1,3,1) + t(0,-6,-2)

    Length AB = |p| = √(6^2 + 2^2) = 2√10

    Unit vector, u = (-3√10/10)j + (-√10/10)k

    q - a = i-2j-2k

    Length AN = u.(q-a) = (4√10)/5

    ∴ Perpendicular distance:

    Length NQ = √(AQ^2-AN^2) = √65/5


    How is this? Yes, I made a mistake with the unit vector components.


    @Joffan:

    I can. Since length AN is (4√10)/5, the coordinates of N would be (1,3,1)+((4√10)/5)(0,-6,-2). I won't simplify here because the exact answer is in rational form.


    By the way, is there a way to write vectors as column vectors here? I don't like writing in component form. Too much work :S.
     
  6. May 18, 2012 #5

    HallsofIvy

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    In LaTeX,
    [ tex ]\ begin{bmatrix}a \\ b \\ c\ end{bmatrix}[ /tex ]
    (without the spaces) gives
    [tex]\begin{bmatrix}a \\ b \\ c\end{bmatrix}[/tex]
     
  7. May 18, 2012 #6
    Your position of N is wrong, because t is not in coordinate units.

    The coordinates of N can be specified without surds.
     
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