Calculate the coordinates of the orthogonal projection of a point on a segment

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Faced with the following task:

Known coordinates 3x points A B C on the plane. The points A B are the beginning and end of the segment AB. You need to find the coordinates of the orthogonal projection (точка D) of the point С on the segment АВ. How do I do this?

Author: Kromster, 2019-01-17
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2 answers

The projection can be found using the scalar product of the vectors

 D = A + AB * Dot(AC, AB) / Dot(AB, AB)

In pseudocode:

 abx = B.X - A.X
 aby = B.Y - A.Y
 dacab = (C.X - A.X) * abx + (C.Y - A.Y) * aby
 dab = abx * abx + aby * aby
 t = dacab / dab
 D.X = A.X + abx * t
 D.Y = A.Y + aby * t

Here slightly different notation (C=P, N=D) and explanations

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 3
Author: MBo, 2019-01-18 04:24:42

  • "Intersection" method:

    If you plot a point С' = (Cx - (By - Ay), Cy + (Bx - Ax)), then the line CC' will be perpendicular to the line AB. The intersection of the lines AB and CC' will give you a point D.

    The method is "heavy" from a computational point of view due to the fact that it uses the intersection of two straight lines as a primitive. But it is easy to implement, if such a primitive is already at hand.

  • Distance method:

    If AA, BB and CC - coefficients equations of a straight line containing a segment AB (easily calculated using the coordinates of the points A and B), then the value

    DD = AA * Cx + BB * Cy + CC

    Will give you the signed distance from this straight line to the point C, multiplied by |(AA, BB)|. If we add the vector DD * (-AA, -BB) to the point C, we get a point that is shifted relative to C in the right direction, but, conditionally speaking, "too far": the distance exceeds the required |(AA, BB)|^2 times. It is enough to divide the offset by this value - we get the required point D

                 DD 
D = C + ------------ * (-AA, -BB)
        |(AA, BB)|^2

    This method is also somewhat cumbersome due to the "extra" calculations of the equation of a straight line, but it can be useful if you already know/calculate the equation of a straight line, and also if you need to calculate the distance from C to AB in addition to the projection point.

  • "Scalar product" method:

    It is obvious that 

            |(A, D)|
D = A + -------- * (A, B)
        |(A, B)|

    In this case, the scalar product of the vector (A, C) by the vector (A, B) is the length of the projection (A, D), multiplied by |(A, B)|. 

    (A, C)•(A, B) = |(A, D)| * |(A, B)|

    Then

            (A, C)•(A, B)
D = A + ------------- * (A, B)
          |(A, B)|^2

(It can be shown that if you put all the calculations of the second method into one formula, it will "reduce" to the third method.)

 0
Author: AnT, 2019-01-17 23:17:54