Perpendicular on a line segment from a given point

问题:

I want to calculate a point on a given line that is perpendicular from a given point.

I have a line segment AB and have a point C outside line segment. I want to calculate a point D on AB such that CD is perpendicular to AB.

I have to find point D.

It quite similar to this, but I want to consider to Z coordinate also as it does not show up correctly in 3D space.

回答1:

Proof: Point D is on a line CD perpendicular to AB, and of course D belongs to AB. Write down the Dot product of the two vectors CD.AB = 0, and express the fact D belongs to AB as D=A+t(B-A).

We end up with 3 equations:

 Dx=Ax+t(Bx-Ax)
 Dy=Ay+t(By-Ay)
(Dx-Cx)(Bx-Ax)+(Dy-Cy)(By-Ay)=0

Subtitute the first two equations in the third one gives:

(Ax+t(Bx-Ax)-Cx)(Bx-Ax)+(Ay+t(By-Ay)-Cy)(By-Ay)=0

Distributing to solve for t gives:

(Ax-Cx)(Bx-Ax)+t(Bx-Ax)(Bx-Ax)+(Ay-Cy)(By-Ay)+t(By-Ay)(By-Ay)=0

which gives:

t= -[(Ax-Cx)(Bx-Ax)+(Ay-Cy)(By-Ay)]/[(Bx-Ax)^2+(By-Ay)^2]

getting rid of the negative signs:

t=[(Cx-Ax)(Bx-Ax)+(Cy-Ay)(By-Ay)]/[(Bx-Ax)^2+(By-Ay)^2]

Once you have t, you can figure out the coordinates for D from the first two equations.

 Dx=Ax+t(Bx-Ax)
 Dy=Ay+t(By-Ay)

回答2:

function getSpPoint(A,B,C){
    var x1=A.x, y1=A.y, x2=B.x, y2=B.y, x3=C.x, y3=C.y;
    var px = x2-x1, py = y2-y1, dAB = px*px + py*py;
    var u = ((x3 - x1) * px + (y3 - y1) * py) / dAB;
    var x = x1 + u * px, y = y1 + u * py;
    return {x:x, y:y}; //this is D
}

回答3:

There is a simple closed form solution for this (requiring no loops or approximations) using the vector dot product.

Imagine your points as vectors where point A is at the origin (0,0) and all other points are referenced from it (you can easily transform your points to this reference frame by subtracting point A from every point).

In this reference frame point D is simply the vector projection of point C on the vector B which is expressed as:

// Per wikipedia this is more efficient than the standard (A . Bhat) * Bhat
Vector projection = Vector.DotProduct(A, B) / Vector.DotProduct(B, B) * B

The result vector can be transformed back to the original coordinate system by adding point A to it.

回答4:

A point on line AB can be parametrized by:

M(x)=A+x*(B-A), for x real.

You want D=M(x) such that DC and AB are orthogonal:

dot(B-A,C-M(x))=0.

That is: dot(B-A,C-A-x*(B-A))=0, or dot(B-A,C-A)=x*dot(B-A,B-A), giving:

x=dot(B-A,C-A)/dot(B-A,B-A) which is defined unless A=B.

回答5:

What you are trying to do is called vector projection

回答6:

Here i have converted answered code from "cuixiping" to matlab code.

function Pr=getSpPoint(Line,Point)
% getSpPoint(): find Perpendicular on a line segment from a given point
x1=Line(1,1);
y1=Line(1,2);
x2=Line(2,1);
y2=Line(2,1);
x3=Point(1,1);
y3=Point(1,2);

px = x2-x1;
py = y2-y1;
dAB = px*px + py*py;

u = ((x3 - x1) * px + (y3 - y1) * py) / dAB;
x = x1 + u * px;
y = y1 + u * py;

Pr=[x,y];

end

回答7:

Since you're not stating which language you're using, I'll give you a generic answer:

Just have a loop passing through all the points in your AB segment, "draw a segment" to C from them, get the distance from C to D and from A to D, and apply pithagoras theorem. If AD^2 + CD^2 = AC^2, then you've found your point.

Also, you can optimize your code by starting the loop by the shortest side (considering AD and BD sides), since you'll find that point earlier.

回答8:

I didn't see this answer offered, but Ron Warholic had a great suggestion with the Vector Projection. ACD is merely a right triangle.

1. Create the vector AC i.e (Cx - Ax, Cy - Ay)
2. Create the Vector AB i.e (Bx - Ax, By - Ay)
3. Dot product of AC and AB is equal to the cosine of the angle between the vectors. i.e cos(theta) = ACx*ABx + ACy*ABy.
4. Length of a vector is sqrt(x*x + y*y)
5. Length of AD = cos(theta)*length(AC)
6. Normalize AB i.e (ABx/length(AB), ABy/length(AB))
7. D = A + NAB*length(AD)

回答9:

Here is a python implementation based on Corey Ogburn's answer from this thread.
It projects the point q onto the line segment defined by p1 and p2 resulting in the point r.
It will return null if r falls outside of line segment:

def is_point_on_line(p1, p2, q):

    if (p1[0] == p2[0]) and (p1[1] == p2[1]):
        p1[0] -= 0.00001

    U = ((q[0] - p1[0]) * (p2[0] - p1[0])) + ((q[1] - p1[1]) * (p2[1] - p1[1]))
    Udenom = math.pow(p2[0] - p1[0], 2) + math.pow(p2[1] - p1[1], 2)
    U /= Udenom

    r = [0, 0]
    r[0] = p1[0] + (U * (p2[0] - p1[0]))
    r[1] = p1[1] + (U * (p2[1] - p1[1]))

    minx = min(p1[0], p2[0])
    maxx = max(p1[0], p2[0])
    miny = min(p1[1], p2[1])
    maxy = max(p1[1], p2[1])

    is_valid = (minx <= r[0] <= maxx) and (miny <= r[1] <= maxy)

    if is_valid:
        return r
    else:
        return None