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Distance between a point and plane

  1. Mar 31, 2016 #1
    • Member warned that template must be used
    Conventionally we have the following equation of a plane ax+by+cz=d where d = ax0+by0+cz0. Where (x0,y0,z0) is a known point on the plane.

    Now if we try to find the distance between a point P and a plane we take any point on the plane Q (x,y,z) and find the vector from Q to P and project on the normal vector. Problem is... when we put the co-ordinates of point P to evaluate "d" in the equation of plane. In the equation of plane, (x0,y0,z0) in "d" is of a point that is on the plane. But in this case P is outside the plane.
    Am I missing out on something here?

    For example please see: http://mathinsight.org/distance_point_plane_examples
     
  2. jcsd
  3. Mar 31, 2016 #2

    Mark44

    Staff: Mentor

    If P is a point that is not on the plane, you can't use it to find d, the constant in the plane equation.
     
  4. Mar 31, 2016 #3
    Hi SamitC:

    I don't understand what the above means. Why do you think P is used to "evaluate" d? Since the equation of the plane
    ax+by+cz=d​
    is known, a, b, c, and d must also be known. What do you mean by "evaluate", since "evaluate d" can't mean "solve for d". My guess is that the problem wants you to calculate the closest distance between P and plane. That means you need to find a point Q = (x,y,z) in the plane where the vector P-Q is normal to the plane. Then the length of P-Q is the desired distance.

    Hope this helps.

    Regards,
    Buzz
     
    Last edited: Mar 31, 2016
  5. Mar 31, 2016 #4

    Mark44

    Staff: Mentor

    I'm not the OP. Please direct your comments to SamitC, not me.
     
  6. Mar 31, 2016 #5
    Hi Mark:

    I apologize for being careless in writing the addressee's name. Thanks for pointing my error out to me.

    Regards,
    Buzz
     
  7. Mar 31, 2016 #6

    Ssnow

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    @SamitC in order to understand well your question if it possible to clarify what are the data of the problem that you have, because if you want to find the distance between a point and the plane, the essential data can be the point ##P## and the equation of the plane ##ax+by+cz+d=0##
     
  8. Mar 31, 2016 #7
    Hi Buzz,
    Thanks for your reply. Let me try to resolve it as per your direction.
    Suppose the point P is (4, - 4, 3). Plane is 2x-2y+5x+8=0. Point Q = (a,b,c).
    Vector QP = <4 - a, - 4 - b, 3 - c>.
    Now this is a scalar multiple of the normal vector n = <2, - 2, 5> that we can get from the equation of the plane. We have so many unknowns. Still I do not know how to calculate the distance?

    I am not able to use the "8" in the equation of the plane anywhere. Can you pls. help me with this confusion.
    Regards,
     
    Last edited: Mar 31, 2016
  9. Mar 31, 2016 #8
    Hi Ssnow,
    Thanks for your reply. Let us take the example: I have a plane 2x-2y+5x+8=0 and point (4,-4,3). We need to find the distance.
    I have worked it out to whatever extent i could... I have uploaded in picture format. I know I am missing something but can't figure out. I am not using the "8" in the equation of the plane. That holds vital information. But I am not sure how to make use of it.

    Can you help me find the distance in this case?
    Regards,
     

    Attached Files:

  10. Mar 31, 2016 #9
     
    Last edited: Mar 31, 2016
  11. Mar 31, 2016 #10
     
    Last edited: Mar 31, 2016
  12. Mar 31, 2016 #11

    LCKurtz

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    I corrected your typo. Let's call the point ##P(4,-4,3)##. This is just some point, not having anything to do with the plane. You want to know how far from the plane it is. In your example, you now look for a point on the plane. This is always real easy. You can see ##Q(0,4,0)## works by inspection and the 8 is used at that step. Now, as your OP example suggests, you just calculate ##\vec V = P-Q## to get a vector across from the plane to the point ##P##. So in this example you have ##\vec V = \langle 4,-8,3\rangle##. You already have a normal vector ##\vec N = \langle 2,-2,5\rangle## which you can make a unit vector ##\hat N## by dividing by its length ##\sqrt {33}##. Then the distance is just the magnitude of the component of ##\vec V## in the direction of ##\hat N##:$$
    d = \left|\vec V \cdot \hat N\right| $$Using the absolute values means you don't have to worry about which side of the plane the point ##P## is on or which side the normal ##\vec N## points.

    To convince your self that this all works, do the problem again with a different point chosen for ##Q## and compare.
     
  13. Mar 31, 2016 #12

    Ssnow

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    Another procedure in order to find the distance is the following (less or more is the same of the previous). You can start to write the line in parametric form from ##P## that is perpendicular to the plane, this is ## r= (4+2t,-4-2t,3+5t)##. You intersect this with the plane and you will find the point in the intersection ##Q##. After you can obtain the distance using the distance formula between two points ##P## and ##Q##.
     
  14. Mar 31, 2016 #13
    Now I am clear with this. Not sure why I did not understand this earlier.
    In the above equation of the plane we have information about a specific point on the plane which is d. We need to utilize this information by using this point in calculating the distance. Earlier I was taking Q as any point. Instead, Q should be this specific point that is in "d".

    Thank you all for your help.
    Regards,
    Samit
     
  15. Mar 31, 2016 #14
    Thank you very much. Once I linked Q to "d" I am clear now.
    Regards
     
  16. Mar 31, 2016 #15

    HallsofIvy

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    A crucial point in this and in similar problems is that "the distance from set A to set B" is defined as the shortest of all distances from a point in set A to a point in set B. In the case of a point and a line, it is easy to show that this is along the line through the given point perpendicular to the given plane. Here, in your example, the plane is given by 2x-2y+5x+8=0. A vector perpendicular to that plane is <2, -2, 5> and a line in that direction, through the point (4,-4,3) is x= 4+ 2t, y= -4- 2t, z= 3+ 5t. Determine where that line crosses the plane, by setting those formulas for x, y, z to get 4(4+ 2t)- 2(-4- 2t)+ 5(3+ 5t)= 0. Solve that for t, use that to determine the point and find the distance between that point and (4, -4, 3).
     
  17. Mar 31, 2016 #16

    LCKurtz

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    I'm afraid that doesn't make any sense at all. "d" is not a specific point in the plane and it doesn't make sense to say "Q should be this specific point that is in "d"." When you have the equation of a plane given, such as ##2x-2y+5z=-8##, it describes a plane in the sense that the set of points ##(a,b,c)## that satisfy the equation form a plane. To find a point in the plane, you just have to pick three numbers ##a,b,c## which satisfy the equation of the plane.
     
  18. Mar 31, 2016 #17
    Hi SamitC:

    What is the known point (x0,y0,z0)?

    ADDED

    You said:
    In the above equation of the plane we have information about a specific point on the plane which is d.​

    d is NOT a point in the plane. (x0,y0,z0) is a point in the plain. d is the DOT product of (x0,y0,z0) and (a,b,c).

    Do you know about DOT products?

    Regards,
    Buzz
     
  19. Mar 31, 2016 #18
    Hi LCKurt:

    I disagree that N=⟨2,−2,5⟩ is a vector normal to the plane. N is the vector from the origin (0,0,0) point and the N point (2,-2,5).
    The point in the plane closest to N is Q= (x0, y0,z0).

    Regards,
    Buzz
     
  20. Mar 31, 2016 #19

    LCKurtz

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    We were talking about the plane ##2x-2y+5z=-8##. ##\vec N =\langle 2,-2,5\rangle ## most certainly is perpendicular to that plane. You can also think of it as a position vector from the origin to the point ##(2,-2,5)## if you want to, but that point is irrelevant to the problem.
     
  21. Mar 31, 2016 #20
    Hi LCKurtz:

    I apologize for my senior moment stupidity. Of course N is normal to the plane. I think my aberration was that for some reason I had it in my mind that the origin (0,0,0) was also in the plane.

    Regards,
    Buzz
     
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