Hello! How to define on which plane lies a given point with coordinates? I have coordinates of plane and normal

Hello everyone. I undertsand it might be very simple question, but I could not solve it. I have point for example A(1.5, 1.5, 3.0) and i have 6 planes of cube and I need to determine on which plane lies this point. I found normal for each plane and need to find for which of these planes ( number of plane) rely A.

Thank you so much in advance for your time and consideration . More appreciate help and any advice.

X0 = 1.5;

Y0 = 1.5;

Z0 = 3.0;

planes(:,:,1) = [0 3 3; 0 0 3; 0 3 0; 0 0 0; 0 0 0];

planes(:,:,2) = [0 0 3; 3 0 3; 0 0 0; 3 0 0; 0 0 0];

planes(:,:,3) = [3 0 3; 3 3 3; 3 0 0; 3 3 0; 3 0 0];

planes(:,:,4) = [3 3 3; 0 3 3; 3 3 0; 0 3 0; 0 3 3];

planes(:,:,5) = [0 3 0; 3 3 0; 0 0 0; 3 0 0; 0 0 0];

planes(:,:,6) = [0 3 3; 3 3 3; 0 0 3; 3 0 3; 0 0 3];

location_plane = 6;

for j=1:6 % j is number of plane

j

plane = planes(:,:,j);

p0 = plane(1,:); %p0 is top left point of plane

p1 = plane(2,:); %p1 is top right point of plane

p2 = plane(3,:); %p2 is bottom left point of plane

p3 = plane(4,:); %p3 is bottom right point of plane

V0 = plane(5,:); %point on the plane

% Pi is initial start point on the ray

Pi = [X0 Y0 Z0]; %initial start point

%Ri = [XBar YBar ZBar]; %direction vector with unit length

A = p0-p2; %calculate A and B then

B = p0-p3; %then to calculate Normal of each plane

n=cross(A,B); % Normal for each Plane

n

Answer 1

[John D'Errico]

Does a point lie in a plane? This is quite easy to do.

The equation of a plane is a simple one. In fact, it works in any number of dimensions. I'll write it out using the dot function, to make it clear that it uses a dot product. If P0 is any point in the plane, and N is the normal vector to the plane, then the equation that tells if X lies in the plane is simply:

dot(X - P0,N) == 0

Of course, you NEVER want to test if something is exactly zero in floating point arithmetic, so you might do this instead:

abs(dot(X - P0,N)) < tol

where tol is some sufficiently small number.

First, you need to start working with VECTORS. I see you did that later on, but get used it it.

XYZ = [1.5, 1.5, 3]; % a point to test

planes(:,:,1) = [0 3 3; 0 0 3; 0 3 0; 0 0 0; 0 0 0];
planes(:,:,2) = [0 0 3; 3 0 3; 0 0 0; 3 0 0; 0 0 0];
planes(:,:,3) = [3 0 3; 3 3 3; 3 0 0; 3 3 0; 3 0 0];
planes(:,:,4) = [3 3 3; 0 3 3; 3 3 0; 0 3 0; 0 3 3];
planes(:,:,5) = [0 3 0; 3 3 0; 0 0 0; 3 0 0; 0 0 0];
planes(:,:,6) = [0 3 3; 3 3 3; 0 0 3; 3 0 3; 0 0 3];

% a point in each plane. I'll just use the first point in each plane
% I could also have used squeeze here
PiP = reshape(planes(1,:,:),3,6)';

% normal vectors to each plane
Normals = zeros(6,3);
for i = 1:6;
  % subtract off that first point, then apply null
  % the result will be a column vector, so transpose it at the end.
  Normals(i,:) = null(planes(:,:,i) - PiP(i,:));
end
PiP

PiP = 6×3

0 3 3 0 0 3 3 0 3 3 3 3 0 3 0 0 3 3

Normals

Normals = 6×3

1 0 0 0 -1 0 1 0 0 0 1 0 0 0 1 0 0 1

Now, is the point XYZ in any of those planes? We could get fancy for ALL of the planes, and do the test in one line of code. For example:

dot(XYZ - PiP,Normals,2)

ans = 6×1

1.5000 -1.5000 -1.5000 -1.5000 3.0000 0

And that suggests it is plane number 6 that we want to find. But a tolerance makes sense still. So you should do this:

tol = eps*100; % a reasonably small number
find(abs(dot(XYZ - PiP,Normals,2)) < tol)

ans = 6

We could have done the same thing using a loop of course.

tol = eps*100;
for i = 1:6
  if abs(dot(XYZ - PiP(i,:),Normals(i,:))) < tol
    loc = i;
    disp("plane # "+loc+" was identified")
    break
  end
end

plane # 6 was identified

As you can see, nothing more than the equation of a plane was ever needed to determine if the point lies in one of those planes. Be careful to make sure the tolerance is not too tight, but 100*eps is fine for this. I suppose, if you wanted to be a little better about it, I might have used a tolerance that would depend on the numbers themselves. So perhaps this:

smartertol = std(PiP(:))*10*eps

smartertol = 3.3415e-15

Note that if your numbers scale in various ways, so the cube gets larger or smaller, then smartertol also scales with them nicely. As well, we could even have been tricky and computed the normal vectors without a loop, using the pagesvd function, but that starts to get a little (unecessarily) deep for this answer.

[Aknur]

Hello @John D'Errico thank you for your help and clear explanation about do not use determinants.

This code works for me very well. But one question whats is loc = i;

Thank you in advance

[John D'Errico]

It (loc) tells which plane was identified.

Did you not see the next line?

disp("plane # "+loc+" was identified")

what does that do?

[Aknur]

@John D'Errico my apologize, I did not see the next line

Answer 2

[Davide Masiello]

Four points can be verified to belong on the same plane if the determinant of the matrix

is equal to zero.

I'd translate that into code in the following way.

x0 = 1.5;      
y0 = 1.5;
z0 = 3;
    
planes(:,:,1) = [0 3 3; 0 0 3; 0 3 0; 0 0 0; 0 0 0];
planes(:,:,2) = [0 0 3; 3 0 3; 0 0 0; 3 0 0; 0 0 0];
planes(:,:,3) = [3 0 3; 3 3 3; 3 0 0; 3 3 0; 3 0 0];
planes(:,:,4) = [3 3 3; 0 3 3; 3 3 0; 0 3 0; 0 3 3];
planes(:,:,5) = [0 3 0; 3 3 0; 0 0 0; 3 0 0; 0 0 0];
planes(:,:,6) = [0 3 3; 3 3 3; 0 0 3; 3 0 3; 0 0 3];

for j = 1:6
    x(j,:) = planes(1:3,1,j);
    y(j,:) = planes(1:3,2,j);
    z(j,:) = planes(1:3,3,j);
end
for j = 1:6
    M = [x(j,:),x0;y(j,:),y0;z(j,:),z0;ones(1,4)];
    if det(M) == 0
        fprintf('The point belongs to face %d.\n',j)
        break
    end
end

The point belongs to face 6.

Examples with other points

x0 = 1.5;      
y0 = 3;
z0 = 1;
for j = 1:6
    M = [x(j,:),x0;y(j,:),y0;z(j,:),z0;ones(1,4)];
    if det(M) == 0
        fprintf('The point belongs to face %d.\n',j)
        break
    end
end

The point belongs to face 4.

x0 = 1.5;      
y0 = 0;
z0 = 2;
for j = 1:6
    M = [x(j,:),x0;y(j,:),y0;z(j,:),z0;ones(1,4)];
    if det(M) == 0
        fprintf('The point belongs to face %d.\n',j)
        break
    end
end

The point belongs to face 2.

Only thing, for edges 2 answers would be correct, but this code will output only the first of the 2 faces the point belongs to.

[John D'Errico]

UGH! Please don't teach people to use a determinant to test for this! Using a determinant is essentially NEVER a good thing to do.

And testing for ANY number to be zero in floating point arithmetic is an insane thing to do!!!!!!!!! I'm sorry, but this is just the wrong advice to offer. Determinants are thigns we see in classes, usually where beginnign linear algebra is taught. They are neat things to teach a student, or so I assume. It is a nice easy ttest to show a student to see if a matrix is singular. How could that be a bad idea? The problem is, when a student computes the determinant of an entirely integer matrix, they get an integer result. But computers use floating point arithmetic, not integer arithmetic. And they don't use the same method you used as a student to compute that determinant. The result is, when you compute the determant of even entirely integer matrix, the result will not even be an integer. For exmple:

format long g
A = magic(4) %Purely integer

A = 4×4

16 2 3 13 5 11 10 8 9 7 6 12 4 14 15 1

rank(A) % and a singular matrix, so det should be zero, right?

ans = 
     3

det(A)

ans = 
      -1.4495071809506e-12

det(A) == 0

ans = logical
   0

So here, A is an entirely integer matrix, that is known to be singular (as a even order magic mattrix.) It must have a zero determinant, right? NOPE. Not exactly zero.

Worse, determinants don't cale well at all. AN identityy matrix is NEVER singular, right? ANd any multiple of an identity matrix is just as well condeitioned, right?

E = eye(10);
det(E)

ans = 
     1

det(E*1000)

ans = 
                     1e+30

det(E/1000)

ans = 
                     1e-30

Is E/1000 singular? That sure seems like a small number. In fact, if E was much larger, the determinant would underflow, and det would tell you the determinant is EXACTLY zero.

Teachers taught you to use determaninats, because their teachers taught them that. They do it because their books show the use of determinants. And because your teachers know only what they have been taught. But that does not make the use of determinants to be a good thing. And your teachers were wrong.

Yes, it is arguably ok to teach about determinants, as everybody will see them eventually, but what they needed to do was to teach another lesson, why you should NOT be using them!!!!!!

SO PLEASE DON"T KEEP THE CHAIN UNBROKEN. LEARN NOT TO TEACH STUDENTS TO DO SOMETHING THAT IS JUST A BAD IDEA. If you go on teaching the same things, then yet another student will learn the wrong lessons, and so they will learn to teach their students one day to use determinants. And that would be a pity.

Perhaps the worst thing is, you never needed to use a determinant at all for this!!!!!!

[Davide Masiello]

Hi @John D'Errico, thanks for the good advice. I guess I didn't think about the more profound numerical implications of using a determinat. I'll keep that in mind for the future :)

[Aknur]

Hi @Davide Masiello thank you for your advice, it works perfect. Thank you for this code and help, it helps me to understand the logic how to find if point lies on a plane.

[Davide Masiello]

@Aknur please do consider @John D'Errico's comments about the limitations of using determinants and equality to zero.

[Aknur]

Hello @Davide Masiello yes I see. Thank you for reminder