Create plane from 3 points and distance from plane to point

[Mar10n]

I have some trouble getting this small problem right

I want to make a plane from these three points:
1
5.004.218 993.185 9.591

2
5.005.202 993.397 12.622

3
5.006.043 993.715 10.325


I have made two vectors from these points
1-2
984 212 3.031

1-3
1.825 530 734

and then calculated the normal vector to these two vectors

n
-1450822 4809319 134620

and from this I have tried to get the equation to the plane which the three lays onto


plane
-1450822 4809319 134620 -2,48239E+12


I want to calculate the distance from the plane to a fourth point

4
5.020.399 933.806 9.908


I have the made this

7,28371E+12 4,49097E+12 1333814960 2,48239E+12

and this to get the distance

(7,28371E+12 + 4,49097E+12 + 4,49097E+12 + 2,48239E+12)/ sqroot(-1450822^2 + 4809319^2 + 134620^2) = 2837385,029

Can someone please have a look at my calculations so far, and tell me what I have done wrong? (I know the answer is wrong...)

I have made it in a excel spreadsheet if someone would look at it

 

[soandos]

not sure if this is how you are supposed to do it, but if you take the circle that intercepts all three points, and then expand the radius to oo, and that will define a plane, albeit in a different way that normal.

 

[Mar10n]

I guess I could do it that way, but I don't think it makes it any easier to calculate the distance from the circle to the fourth point?

 

[Tac-Tics]

That seems like it would make things very complicated.

To solve something like this, you have your three points x, y, z. Take the vectors x-y and x-z and find a vector mutually perpendicular to both with the cross product: n = (x-y) X (x-z) This is my vector normal to x-y and x-z:n=-1450822i 4809319j 134620k. This is the normal to your plane. It helps to have it of unit length, so take n' = n / |n| Sorry, I don't understand this part what is n / |n| ?.

All points on your plane satisfy the plane equation. That is if p0 is a point on your plane, a point p is also on your plane if and only if

p0 * n' = p * n

Where * is the dot product.

Take any point w. From w, the shortest path to the point is the straight line passing through w parallel to the normal. This line is given by {w + a n' | for all a}. To find the closest point on the plane, you must simply find for which value of a w + a n' is on the plane. That means we use the plane equation: (w + a n') * n' = p0 * n'. Solve for a.

What is the distance from w to the plane? After you've found a, it's simply |w - w + a n'| = |a n'|, and since n' has unit length, the distance reduces to |a|.

Now you just need to plug & chug!

EDIT: I'm a little bit retarded I think. I edited this post instead of quoting it!

 

[Mar10n]

My comments is in red

That seems like it would make things very complicated.

To solve something like this, you have your three points x, y, z. Take the vectors x-y and x-z and find a vector mutually perpendicular to both with the cross product: n = (x-y) X (x-z) This is my vector normal to x-y and x-z:n=-1450822i 4809319j 134620k. This is the normal to your plane. It helps to have it of unit length, so take n' = n / |n| Sorry, I don't understand this part what is n / |n| ?.

All points on your plane satisfy the plane equation. That is if p0 is a point on your plane, a point p is also on your plane if and only if

p0 * n' = p * n

Where * is the dot product.

Take any point w. From w, the shortest path to the point is the straight line passing through w parallel to the normal. This line is given by {w + a n' | for all a}. To find the closest point on the plane, you must simply find for which value of a w + a n' is on the plane. That means we use the plane equation: (w + a n') * n' = p0 * n'. Solve for a.

What is the distance from w to the plane? After you've found a, it's simply |w - w + a n'| = |a n'|, and since n' has unit length, the distance reduces to |a|.

Now you just need to plug & chug!

 

[Tac-Tics]

To solve something like this, you have your three points x, y, z. Take the vectors x-y and x-z and find a vector mutually perpendicular to both with the cross product: n = (x-y) X (x-z). This is my vector normal to x-y and x-z:n=-1450822i 4809319j 134620k

Yeah. It looked like you did it correctly up to this point. Whether or not the numbers are right, I don't know =-)

This is the normal to your plane. It helps to have it of unit length, so take n' = n / |n|.
Sorry, I don't understand this part what is n / |n| ?

Sorry about that. |n| would be the length of a vector n. So n / |n| would be the vector in the same direction as n, but with unit length. Converting a vector to unit length is somewhat confusingly called "normalizing" it. It's like the "normal" in "orthonormal" and completely different from the "normal" in "normal to a plane"!