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Find shortest distance from a point along a vector to a plane

  1. Feb 4, 2012 #1
    1. The problem statement, all variables and given/known data

    The vector from the origin, O, to point P has magnitude 60 m and has equal direction angles with the x, y, and z axes. Find the shortest distance from point P to the plane containing points A, B, and C.

    A (12,0,0)
    B (0,16,0)
    C (0,0,9)

    2. Relevant equations

    N/A

    3. The attempt at a solution


    My initial thoughts were that the angle w/ respect to the x, y, and z axes must each be 45 degrees (in order to have equal direction angles).

    I tried solving by taking point P to have coordinates (42.43, 42.43, 42.43) because 60*cos(45) and 60*sin(45) equal 42.43 then taking x = 42.43t, y = 42.43t, and z = 42.43t as parametric equations and the equation of the plane as 12x+16y+9z=144 but t came out unfathomably low (around 3.5).

    I don't think this is that difficult but I find myself confused.

    Edit: I apologize if this doesn't warrant placement in the physics sub-forum, it's a review question I have for Statics so I had it in my head it's physics. I realize it may be more geometry than anything else.
     
  2. jcsd
  3. Feb 4, 2012 #2

    BruceW

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    To start with, this isn't correct. This would be correct in 2d (polar coordinates). But in 3d, you would have to use spherical polar coordinates. Alternatively, you can just solve without using angles, since you know each component is the same, and you require the sum of their squares to equal the square of the displacement from the origin.
     
  4. Feb 5, 2012 #3
    OK, I understand the 3-D bit. But let me see if I follow you,

    The sum of the squares,

    x2 + x2 + x2 = 3x2

    and the square of the displacement from the origin,

    602 = 3600

    Now,

    3600 = 3x2
    [itex]\frac{3600}{3}[/itex] = x2
    x = 34.64

    But where does the coordinates of the plane come in?
     
  5. Feb 5, 2012 #4

    BruceW

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    You've now got the position of the point, which was the first thing to do. I think there are a few ways to find the distance from the point to the plane. The way I know is by using some of the properties of vectors. To start with, what do we know about the vector from the point to the plane? (just imagine the situation, and you should realise the important property that it has).
     
  6. Feb 5, 2012 #5
    I want to say that the vector is perpendicular to the plane would I don't think that's necessarily true. If it were just a point, then the shortest distance would be a perpendicular line from the point to the plane. I know the vector has to have equal angles w/ the x, y, and z axes but I can't find where that vector crosses through the plane. Knowing the point it ends at, it would simply be a matter of subtracting the point it crosses the plane at to the point it ends at and finding the magnitude of that vector.
     
  7. Feb 5, 2012 #6

    BruceW

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    Yes, the 'shortest' vector from the point to the plane is perpendicular to the plane. I'm not sure what you meant that the vector must have equal angles... The vector from the origin to the point does have equal angles. But the vector from the point to the plane will not have equal angles.

    You're right that if we could find a position vector for the point in the plane closest to the other point, then we can just subtract the position vectors of these two points from each other, then finding the magnitude would give us the shortest distance.

    So you know that the 'shortest' vector from the point to the plane will be in the direction of the normal of the plane. And you know where the point is, so you can use the equation for a straight line to find where the closest point on the plane is.

    (There are other ways to get to the answer, but since you mentioned finding the point on the plane which is closest, you should do it that way if it makes most sense to you, and this is a valid way of finding the answer).
     
    Last edited: Feb 5, 2012
  8. Feb 5, 2012 #7
    OK, so I took the points A(12,0,0), B(0,16,0), and C(0,0,9) and found vectors AB = -12i + 16j + 0k and BC = 0i -16j + 9k. I crossed those vectors and came up w/ 144i + 108j + 192 k then by choosing point A, found the eq. for the plane, 144x + 108y + 192z = 1728.

    I substituted the coordinates (34,64, 34,64, 34.64) in for x, y, and z respectively and subtracted 1728 then divided by the sqrt of the sum of the squares 144, 108, and 192.

    I arrive at 51.87 m.This seems reasonable.
     
  9. Feb 5, 2012 #8

    BruceW

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    I agree with all of this paragraph. I'm not so sure about the next paragraph though. The method you use in the next paragraph looks like it should give the component of the position vector of the point, in the direction of the normal to the plane (which is not what the question is asking for). And when I try to do your method, I get 58.4, not 51.87. Interestingly, your answer of 51.87 is the correct answer. But I don't understand how you arrived at it... Maybe I interpreted your method in the second paragraph incorrectly.
     
  10. Feb 7, 2012 #9
    Whatever I did worked. I checked w/ someone else who did it also and they had the same answer. Thanks for your help.
     
  11. Feb 7, 2012 #10

    BruceW

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    no worries, glad to be of some help. I come on this site and help so that I don't forget how to do these kinds of questions, but also its nice to know if I've helped someone :)
     
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