Equation of the plane that contains a point

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In summary: Keep in mind that the equation of a plane is Ax+By+Cz+D=0, so you need to add D back in when solving for the equation based on the normal and a point in the plane.In summary, we are trying to find the equation of a plane (Ax + By + Cz + D = 0) that contains three given points (p1 = (4,0,4), p2 = (2,-1,8), p3 = (1, 2, 3)). To do this, we first find the normal vector of the plane by taking the cross product of two vectors formed by subtracting one point from the other two points. After finding the normal vector (-7, -14
  • #1
questionman23
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Homework Statement


Let p1 = (4,0,4), p2 = (2,-1,8) and p3 = (1, 2, 3).

Specify the condition that p = (x,y,z) lies in the plane of p1, p2, p3 (as an equation in x, y, z). Equation of a plane has the form of Ax + By + Cz + D = 0.

(There were some other questions before this about finding the normal to the plane and area of the triangle formed by the 3 points which I have completed).

2. The attempt at a solution
I'm not sure where to start here. The normal vector I have is <s>(-7, -9, -7)</s> (-7, -14, -7) which I think is correct. I think I somehow need to incorporate it into the general equation Ax+By+Cz+D=0, but don't know how.

Any help is very much appreciated.
 
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  • #2
I don't think you have the normal right. How did you get it?
 
  • #3
I took the cross product of (p2-p1) and (p3-p1). I found it using the determinant.
 
  • #4
I did too. But I didn't get the same thing. Either there's a typo in the statement. Possible. Or I made a mistake. Also possible. Or you made a mistake. Also possible. It's hard to tell since you didn't show how you worked the cross product in any detail. If you do that people can suggest where you might have gone wrong. Or I can figure out where I did. I got the -7's ok. I wonder about the -9 in the middle.
 
  • #5
Thanks for the quick response. I just double checked and it seems I was wrong. I'm now getting (-7, -14, -7) for the normal. Is that what you got?

v = (p2-p1) = -2, -1, 4
w = (p3-p1) = -3, 2, -1

v X w = (-7, -14, -7)
I then did the determinant of the matrix.
http://en.wikipedia.org/wiki/Cross_product#Matrix_notation

I found a cross product calculator which also seems to give the answer I got.
http://www.analyzemath.com/vector_calculators/vector_cross_product.html

If you need more detail, I'll be happy to explain myself.
 
  • #6
No, s'ok. That's what I got. Now that you have the normal n right, the equation of the plane is n.(x,y,z)+D=0 (dot product). Put anyone of p1, p2 or p3 in for (x,y,z) to get the value of D. If everything is kosher, you get the same value for any of them.
 
  • #7
Ok, let's see. Using p2 (2,-1,8)
-7*(2) + -14*(-1) + -7*(8) = -56
I get -56 when using p1 and p2, as well.
That's D.

For A, B, and C in the equation, I just use the normal components? So the final equation becomes...
-7x + -14y + -7z - 56 = 0.

If that's it, then thanks a bunch!
 
  • #8
Almost. Your first stuff seems to say -7x-14y-7z=-56. Then you say -7x-14y-7z-56 = 0. They are just a little different. Right? Which one is correct?
 
  • #9
Hmm, so the -56 is -D. So it should be -7x-14y-7z + 56 = 0. Thanks for catching that! :)
 
  • #10
No problem.
 

1. What is the equation of the plane that contains a point?

The equation of a plane in three-dimensional space can be written as Ax + By + Cz + D = 0, where A, B, and C are the coefficients of the x, y, and z variables, respectively, and D is a constant. To determine the specific equation of a plane containing a given point, the coordinates of the point must be substituted into the equation to solve for the value of D.

2. How do you find the coefficients of the equation of a plane?

The coefficients of the equation of a plane can be found by using the coordinates of three non-collinear points on the plane. These points can be used to set up a system of three equations, which can then be solved to find the values of A, B, and C. Alternatively, if the normal vector to the plane is known, the coefficients can be determined using the vector's components.

3. Can a plane be defined by only one point?

No, a plane cannot be defined by only one point. A plane is a two-dimensional surface and therefore requires at least two points to define it. However, a single point can be used to determine the equation of a plane.

4. What is the significance of the constant term in the equation of a plane?

The constant term, D, in the equation of a plane represents the distance of the plane from the origin along the direction of the plane's normal vector. It can be positive, negative, or zero, depending on the position of the plane in relation to the origin.

5. Can the equation of a plane be written in different forms?

Yes, the equation of a plane can be written in different forms, such as the vector form or the parametric form. These forms may be more useful in certain situations, but they all represent the same plane and can be converted into each other.

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