在MATLAB中,已知一个点坐标和一个由三点组成的平面（这三点坐标已知）,求这个点在这个平面上的投影坐标例如,已知点A,B,C,D,怎么求点D在平面ABC上的投影点坐标.还有一点：怎么把这些三维坐

在MATLAB中,已知一个点坐标和一个由三点组成的平面（这三点坐标已知）,求这个点在这个平面上的投影坐标
例如,已知点A,B,C,D,怎么求点D在平面ABC上的投影点坐标.
还有一点：怎么把这些三维坐标换算成平面ABC上的二维坐标
这是想用在一个程序里的,希望高手作答,能用编程语言简单实现
有请高手现身!

You might want to review the linear algebra text book of freshman's course in most college.
First of all, let's make some vectors based on your given points:
b=B-A
c=C-A
d=D-A
, In other words, we simply let A be the original point. Vectors b, c, and d are connecting from A to B, to C, and to D, respectively. We are going to answer your questions based on these vectors.
To check if A, B, and C are not in a straight line, we simply need to check if vector b and c are independent. We use the rank of matrix with verctor b and c to determine:
r = rank([b c])
Here we assume b and c are column vectors. (If b and c are row vectors, r = rank([b;c]) in Matlab) Now, If r=2, vector b and c are independent, in other words, A, B and C are not in one straight line, they can determine a plane.
We assume b and c are independent now. We use the following steps to obtain a basis:
(1) take b as the first vector: v1 = b
(2) get the normal vector of the plane determined by b and c:
v3 = cross(b,c)
(3) get the third vector which is orthogonal to both v1 and v3:
v2 = cross(v3, v1)
*Note: v2 must be on the plane defined by b and c, or equavilently by A, B and C
(4) normalize v1, v2 and v3, we get:
e1 = v1 / norm(v1)
e2 = v2 / norm(v2)
e3 = v3 / norm(v3)
Then, e1, e2 and e3 form a basis of whole 3D spance.
To get the projection of D onto plane ABC, you can get the inner product of d and each vector of the basis :
p1 = dot(d,e1)
p2 = dot(d,e2)
p3 = dot(d,e3)
Then:
projectionOfDOntoABC = p1*e1+p2*e2+p3*e3
Finally, e1 and e2 form a basis of the plane ABC. You can use (p1,p2) as the coordinates with respect to the basis e1,e2 to refer a point on the plane.