how to find a normal vector?

for example, there are 2 points. P0(4,3,2) ,P1(8,5,4) and the vector ->P0P1

I know that In three dimension, there are infinite number of vectors perpendicular to a given vector.

but i know the point which is on the plane.

To use this function, I need to find a normal vector of the plane.

In my case, P1 point wil be the V0 and P1 for this function.

[I,check]=plane_line_intersect(n,V0,P0,P1)
%       n: normal vector of the Plane 
%       V0: any point that belong   s to the Plane 
%       P0: end point 1 of the segment P0P1
%       P1:  end point 2 of the segment P0P1

https://kr.mathworks.com/matlabcentral/fileexchange/17751-straight-line-and-plane-intersection

Answer 1

[Matt J]

In my case, P1 point wil be the V0 and P1 for this function.

You need 3 distinct, non-colinear points in a the plane to calculate its normal. If V0,P0,V1 are such points, then you would do,

normal=cross(P1-P0,V-P0)

[Sierra]

Thanks Matt J

as i said, i know one point which is on a plane. to do cross product i made two arbitrary point.

but when i used the function, it only returns x and y value. all z value is zero

do you have any idea to solve this problem?

Here is my code

intersection_point = [];

for i = 1:length(mean_trajectory_double)

    O = [mean_trajectory_double(i,1),mean_trajectory_double(i,2),mean_trajectory_double(i,3)];
    P = [mean_trajectory_double(i,1)+1,mean_trajectory_double(i,2)+1,mean_trajectory_double(i,3)-1000];
    Q = [mean_trajectory_double(i,1)+2,mean_trajectory_double(i,2)+2,mean_trajectory_double(i,3)+1000];

    OP = Q-O;
    OQ = P-O;
    normalvector = cross(OP,OQ);

    for j = 1:length(RKSI_Arr_33R)
        for k = 1:length(RKSI_Arr_33R(j).Latitude)-1

            [I,check]=plane_line_intersect([normalvector(1) normalvector(2) normalvector(3)]...
                ,[mean_trajectory_double(i,1) mean_trajectory_double(i,2) mean_trajectory_double(i,3)]...
                ,[RKSI_Arr_33R(j).Longitude(k) RKSI_Arr_33R(j).Latitude(k)  RKSI_Arr_33R(j).BAlt(k)]...
                ,[RKSI_Arr_33R(j).Longitude(k+1) RKSI_Arr_33R(j).Latitude(k+1)  RKSI_Arr_33R(j).BAlt(k+1)]);
        end
    end
    intersection_point = [intersection_point;I];
end

O is the point I know already and P, Q is what i made.

mean_trajectory_double(i,1) is longitude

mean_trajectory_double(i,2) is latitude

mean_trajectory_double(i,3) is altitude

I changed the P and Q's altitude several time, but it always returend zero.

and in making two arbitrary point(P,Q), if there are better way, tell me please.

Thanks.

[Torsten]

I don't understand your question.

If P0P1 is perpendicular to the plane and P1 lies in the plane, then the equation of the plane is

(-P0 + P1).' * x = (-P0 + P1).' * P1

where P0, P1 are given as column vectors.

[Sierra]

I know that. What I'm asking is that to get the normal vector, I need 3 points on a plane.

but i know only one point on a plane. so how can i get the other two points?

Thanks Torsten.

[Torsten]

P0 = [4 3 2];
P1 = [8 5 4];
normal_to_plane = (-P0 + P1);
% Generate two vectors that span the plane
in_plane = null(normal_to_plane)
% Generate two points in the plane
Q = P1.' + in_plane(:,1)
S = P1.' + in_plane(:,2)
% Test whether points Q, S are in plane
test_Q_in_plane = normal_to_plane*Q - normal_to_plane*P1.'
test_S_in_plane = normal_to_plane*S - normal_to_plane*P1.'

[Matt J]

but i know only one point on a plane. so how can i get the other two points?

You said you know 3 points in the plane: P0,P1,V0

As I said in my original answer, you must be given 3 non-colinear points in the plane to determine its equaiton. It is a minimum requirement.

If you want to know only its normal, you must be given two vectors parallel to the plane. This is also a minimum requirement.

[Sierra]

All i know is P0 and P1. and I want to know Q and S point.

[Matt J]

There is no relationship between S,Q and P0,P1 conveyed in your picture. Also, your question asked how to determine the normal to the plane. But the difference vector P1-P0 is the normal to the plane, so you already know it. S and Q have nothing to do with anything.

Also, what happened to V0? Where is V0 in your picture?

[Torsten]

I renamed the points in the plane as Q and S in the code above.

[Sierra]

Thanks Torsten, your code worked perfectly.

but I have one problem in 'z' value.

intersection_point = cell(30,3)
        lon = [];
        lat = [];
        alt = [];

for i = 1:length(mean_trajectory_double)-1

    P0 = [mean_trajectory_double(i,:)];
    P1 = [mean_trajectory_double(i+1,:)];
    normal_to_plane = (-P0 + P1);
    P2_in_plane = P1.' + in_plane(:,1);
    P3_in_plane = P1.' + in_plane(:,2);
        lon = [];
        lat = [];
        alt = [];
      
    for j = 1:100
        for k = 1:100

            [I,check]=plane_line_intersect([normal_to_plane(1) normal_to_plane(2) normal_to_plane(3)]...
                ,[mean_trajectory_double(i,1) mean_trajectory_double(i,2) mean_trajectory_double(i,3)]...
                ,[RKSI_Arr_33R(j).Longitude(k) RKSI_Arr_33R(j).Latitude(k)  RKSI_Arr_33R(j).BAlt(k)]...
                ,[RKSI_Arr_33R(j).Longitude(k+1) RKSI_Arr_33R(j).Latitude(k+1)  RKSI_Arr_33R(j).BAlt(k+1)]);
        end
                lon = [lon;I(1)];
                lat = [lat;I(2)];
                alt = [alt;I(3)];
    end
              

    intersection_point{i,1} = [lon];
    intersection_point{i,2} = [lat];
    intersection_point{i,3} = [alt];

end

there is no problem in x(lon),y(lat) value. but z(alt) value print same number.

[Sierra]

To Matt J

I thought P1 is V0, P1 in this function.