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

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!
恐龙出来溜溜 1年前 已收到1个回答 举报

脑震荡的小猪 幼苗

共回答了16个问题采纳率:93.8% 举报

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.

1年前

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