Solving Plane Equation: 3 Coordinates Given

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In short, the distance from the origin to the line is just the length of the component of the vector \vec{B} perpendicular to \vec{A}.In summary, to find the equation of a plane given three points, you can use the cross product of two vectors formed by any two of the points, and then plug in the coordinates of one of the points to solve for the intercept (d) in the plane's equation. To find the angle between a line and a plane, use the dot product between the line's direction vector and the normal vector of the plane. The point of intersection between a line and a plane can be found by solving a system of equations. And to find the shortest distance from a line or plane
  • #1
dcl
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I'm having trouble working out the equation of a plane. Say I'm given 3 coordiantes , how would I work out the equation of the plane in the form of ax + by + cz = d?

say (1,-1,-1), (2,1,2), (2,-2,1) for example.

thank you in advance
 
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  • #2
Let A = (1, -1, -1), B = (2, 1, 2) and C = (2, -2, 1). The vectors A - B and A - C will be parallel to the plane (of course, so will B - C, C - A, and so on). A normal vector to the plane will be given by the cross product between A - B and A - C, let's assume this vector is (n_1, n_2, n_3). The plane will then have the equation n_1x + n_2y + n_3z + d = 0, finding d is simple since you know a point on the plane (in fact, you know three).
 
  • #3
A(1, -1, -1)
B(2, 1, 2)
C(2, -2, 1)

[tex]\vec{AB} = (1, 2, 3)[/tex]
[tex]\vec{BC} = (0, -3, -1)[/tex]

The cross product is:
Code:
    |i  j  k|
n = |1  2  3| = -2i + 9i + 0j + j - 3k - 0k = (7, 1, -3)
    |0 -3 -1|
(At least that's how we were taught to do it.)

So the plane equation is: 7x + y - 3z + D = 0
To find D, substitude the coordinates of one of the points in the equation and solve for D = -9.
 
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  • #4
In 3D (the space with 3 dimensions) the general form of the equation is A*x +B*y +C*z +D=0. You have three points in 3D, which uniquely determine your plane.
Since your points belong to the plane, their coordinates have to verify identically plan’s equation.
Plug in the known values for x, y, and z; you will get a system of 4 equations with four unknowns.
Through elimination’s you should be able to determine A, B, C and D and write the plane's equation, containing your three points in its general form.
Another way is writing a matrix of the system.
Plug in the known values for x, y, and z; you will get a system of 4 equations with 4 unknowns.
You’ll get :
First line: | x y z 1|
2nd line: |x1 y1 z1 1 |
3rd line | x2 y2 z2 1| = |0|
4th line: | x3 y3 z3 1|
This should be equal with zero(if you write it like a matrix, as I tried above). Doing the calculations(elimination or reduction) you should get the plane equation.
If you have problems e-mail me for help.
Here’s a site where you can easily find information regarding general math and good reference:

http://mathworld.wolfram.com/Plane.html
Hope this helps,

Michael
 
  • #5
thank you so much guys for your help :)
 
  • #6
Some more questions.. regarding that plane. :(
We also have a line going through the points (-3.2,1) and (2,1,-5)
I worked out the equation of that line to be
r= (-3,2,1) + t(5,-1,-6)

I have to find the angle the line makes with the plane. The point where they intersect and the shortest distance from the line and the plane to the origin.
 
  • #7
I think you can just find the angle between the line and the normal vector of the plane, using the dot product. And then the angle between the line and the plane is just 90o minus that.
 
  • #8
That makes sense... How do I get the normal vector of the plane?
 
  • #9
If the plane's equation is Ax + By + Cz + D = 0, the normal vector is (A, B, C). We used this eariler, except we found the normal vector and from there proceeded to find the plane's equation.
 
  • #10
The point of intersection of the line and plane can be found by solving the system of equations: Line: [tex](x, y, z) = (k, l, m)t + (a, b, c)[/tex], Plane: [tex]Ax + By + Cz + D = 0[/tex]. So:

[tex]A(kt+a) + B(lt+b) + C(mt+c) = -D[/tex]

Solve for t:

[tex] t = -\frac{D + Aa + Bb + Cc}{Ak + Bl + Cm} [/tex]

And plug it back into the equation for the line to get the point of intersection. (It's easier to do this process in each instance than it is to remember the formula.)

For any point not on a line or plane, the shortest distance is along a line normal to the original line or plane. In particular, for the origin and the plane, the line along which the least distance is accomplished is [tex](A, B, C)t[/tex]. The closest point on the plane to the origin is the point where this line intersects the plane. Finding the point as above gives

[tex] t = -\frac{D}{A^2 + B^2 + C^2}[/tex]

and the point [tex](At, Bt, Ct)[/tex]. The distance from this point to the origin is then [tex]\frac{|D|}{\sqrt{A^2 + B^2 + C^2}}[/tex].

An alternative approach for finding the distance from the line to the origin: Let the line be [tex]\vec{r} = \vec{A}t + \vec{B}[/tex], then you need to minimize [tex]|\vec{r}|[/tex], but the square root is an increasing function, so this is the same as minimizing [tex]r^2 = A^2t^2 + 2\vec{A}\cdot\vec{B}t + B^2[/tex]. The minimum of this quadratic equation in [tex]t[/tex] is found at the vertex: [tex]t = \frac{-\vec{A}\cdot\vec{B}}{A^2}[/tex]. Using this, we find that [tex]r^2 = B^2 - \frac{(\vec{A}\cdot\vec{B})^2}{A^2}[/tex]. Since [tex]\vec{A}\cdot\vec{B} = AB \cos \theta[/tex], where [tex]\theta[/tex] is the angle between them, this simplifies to [tex]r = B\sin \theta[/tex].
 

1. What is the formula for solving a plane equation with 3 coordinates?

The formula for solving a plane equation with 3 coordinates is Ax + By + Cz + D = 0, where A, B, and C are the coefficients of x, y, and z, and D is a constant.

2. How do you find the coefficients of the plane equation with 3 coordinates?

The coefficients of the plane equation can be found by using the coordinates of three non-collinear points on the plane. Plug in the x, y, and z values of these points into the formula Ax + By + Cz + D = 0 and solve for A, B, and C.

3. What is the significance of the constant in the plane equation with 3 coordinates?

The constant, D, in the plane equation represents the distance of the plane from the origin. If D is positive, the plane is in front of the origin; if it is negative, the plane is behind the origin.

4. Can the plane equation with 3 coordinates be used to find the distance between a point and the plane?

Yes, the plane equation can be used to find the distance between a point and the plane by plugging in the coordinates of the point into the equation and solving for the distance.

5. Is there a specific method for solving a plane equation with 3 coordinates?

Yes, there is a specific method called the Gaussian elimination method that can be used to solve a plane equation with 3 coordinates. This method involves using row operations to manipulate the equation until it is in a simplified form.

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