How to measur the inclination angle of a palne?

Let's say I have plane with two known points p1(x1, y1, z1) and p2(x2, y2, z2) somewhere in space. Is there any way to measure the inclination angle\angles for that plane? Which angle from the three angles (with respect to x, y or z axes) can be considered the inclination angle of that plane? Or I need calculate the three inclination angles?

Or I can say it in other way to add more details:

I have four known extrimeties points of a plane. I derived the plain equation from three known points only. Then, I measure the longest diagonal of the plane and I need to measure the inclination angle from that diagonal vector which has two known points. I attached a file that explain the problem with more details.

Can you help me please because i can't find any answer for a while.

Answer 1

[John D'Errico]

Is there a way?

NO. Sorry, but absolutely not since two points do not determine a plane. There are infinitely many planes that will pass through any two points, Each of which can lie at any angle you so desire.

[Susan Smith]

Thank you John for your answer. I updated my question to add more details.

[John D'Errico]

It looks like your question is morphing into something new. Is it solvable? Got me. Despite the vague pdf file you have attached, you are still not explaining the problem.

Your question still seems to revolves around:

"...It is required to find the inclination angle of the vector AB..."

Perhaps the reason why you have gotten no better help than what I said is because you are still posing a question that has no answer.

You say that you have measured the "longest diagonal of the plane" which is meaningless. Sorry, but it is. A plane has no "diagonal". It is apparently just the longest diagonal of a quadrilateral region that you have chosen.

The line AB lies in the plane. But the plane itself can be at any inclination angle despite the presence of a line in the plane. Anyway, inclination angle is not a meaningful concept, UNLESS you tell us to what the angle is against.

That is, you can think of two lines, coming together at a point. You can now talk about the angle between those lines. No problem there.

But given only a plane in space, or worse, a line? On their own, there is no angle. There is nothing to compute.

So what you need to do is explain, more clearly. inclination of what, RELATIVE TO WHAT?

Let me guess. MAYBE you want to know the inclination of the plane, relative to the x-y plane. This you should be able to do, but it will likely have nothing to do with the line AB, since the line AB is just one arbitrary line in the plane.

Or, perhaps, you are asking to find the inclination of that line AB, again perhaps to the (x,y) plane. As long as the line AB will intersect with the (x,y) plane, then you could determine the inclination of that line to the (x,y) plane.

Or, perhaps you want to know the angle the line AB makes against the axis vectors themselves, or, god knows what.

My point is, an inclination is an angle. An angle is something that requires two lines. So far, you have done nothing more than talk about ONE line, though you have gone into great detail describing the line. It seems you are missing the forest for the trees.

[Susan Smith]

Thank you John! Ok, let say the plane is a leaf of a plant and I need to measure the inclination angle with respect to the tree trunk which is the z axis here.

[Susan Smith]

I measured the direction angles of vector OA with respect to the x, y and z axes and I did the same thing for vector OB, but I think this is not correct. Then, I try to subtract point A from point B and then I found the direction angles for the new vector with respect to x, y and z axes. But still not sure if this is true or not.

[John D'Errico]

We are making some progress now.

Assume you have three points on the surface of a leaf. We can assume this is a planar leaf. (Its a pretty boring leaf, if so, but assume that.)

How can we define a plane? The trick is not to think about a plane in terms of a vector that lies in the plane, because that does not define the plane uniquely. Those three points in the plane can be used to determine the vector normal to the plane. This is far more important. Thus if you have three points, A,B,and C in the plane, then the normal vector is:

N = cross(A-C,B-C);

so the vector cross product between the vectors A-C and B-C. The nice thing is, MATLAB already provides a function called cross that does exactly what you need - compute a cross product.

Now, what vector are we computing the angle with? The Z axis vector, so we have a nicely vertical tree. (Gosh, what a boring forest we have. Planar leaves and vertical trees.) But what vector is the z axis?

TreeVec = [0 0 1];

The trick here is a useful cosine formula. The angle between two vectors u and v is given by

cos(theta) = dot(u,v)/(norm(u)*norm(v))

We can find that formula here:

https://en.wikipedia.org/wiki/Cosine_similarity

So how does this apply to your problem? Just solve for theta in that formula.

theta = acos(dot(N,TreeVec)/norm(N)/norm(treeVec));

Theta there will be in radians, since I used acos, not acosd. If you want degrees, use acosd instead.

thetad = acosd(dot(N,TreeVec)/norm(N)/norm(treeVec));

But that is the angle between the two vectors. If you really want the angle between the plane and the vertical tree, you will need to subtract that angle from 90 degrees (or pi/2 radians, depending on how you compute it.)

[Susan Smith]

Ok, thank u very much for your detailed answer. Yes I calculated the plane norm already but is the below equation correct?

thetad = acosd(dot(N,TreeVec)/norm(N) / norm(treeVec));

or do you mean:

thetad = acosd(dot(N,TreeVec)/norm(N) * norm(treeVec)); as the equation above:

cos(theta) = dot(u,v)/(norm(u)*norm(v))

And could you please give a referance for this equation.

Many thanks in advance.