Equation of plane given points

In summary: That is how it is worded, reading it word for word. I thought it was weird too given 6 points instead of the common 3.
  • #1
ParoXsitiC
58
0

Homework Statement



Find equation of plane containing points: (1,1,5),(3,5,3),(8,8,1),(10,2,2),(18,6,-1),(-1,-3,6)


Homework Equations



Find 2 vectors given 3 points, using a common point. The cross product of these 2 vectors will be the normal vector of the plane. Use normal vector coords <a,b,c> as coefficients in the ax+by+cz=d formula where x,y,z is any point in the plane and solve for d. This equation better be true for all points.


The Attempt at a Solution



P = (1,1,5)
Q = (3,5,3)
R = (8,8,1)

PQ = <2,4,-2>
PR = <7,7,-4>

PQ x PR = <-2,-6,14>

Plug in point P into formula and get:

-2x-6y+14z = 62

Test formula by plugging in Q and don't get 62.
 
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  • #2
Check the z-component of your cross product result.
 
  • #3
aftershock said:
Check the z-component of your cross product result.

Okay I had 14 when should be -14

PQ x PR = <-2,-6,-14>

Plug in point P into formula and get:

-2x-6y+14z = -78

Now point Q works out to be -78 but the point (10,2,2) turns out to be -60

Is this a trick question?
 
  • #4
ParoXsitiC said:
Okay I had 14 when should be -14

PQ x PR = <-2,-6,-14>

Plug in point P into formula and get:

-2x-6y+14z = -78

Now point Q works out to be -78 but the point (10,2,2) turns out to be -60

Is this a trick question?


I just used point p and the normal vector we now agree on to form the equation of a plane and got something different than what you have. You might have messed up the algebra.
 
  • #5
aftershock said:
I just used point p and the normal vector we now agree on to form the equation of a plane and got something different than what you have. You might have messed up the algebra.

I forgot to update the formula with -14:


-2x-6y-14z = -78

Still when using (10,2,2) I get -60 and not -78
 
  • #6
ParoXsitiC said:
I forgot to update the formula with -14:


-2x-6y-14z = -78

Still when using (10,2,2) I get -60 and not -78

Yeah that is weird, are you sure they mean all the points are in the same plane?
 
  • #7
aftershock said:
Yeah that is weird, are you sure they mean all the points are in the same plane?

That is how it is worded, reading it word for word. I thought it was weird too given 6 points instead of the common 3
 

FAQ: Equation of plane given points

What is the equation of a plane given three points?

The equation of a plane in three-dimensional space can be written in the form 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 find the equation of a plane given three points, we can use the cross product of two vectors formed by the three points to determine the values of A, B, and C, and then plug in one of the points to solve for D.

What is the significance of the coefficients in the equation of a plane?

The coefficients A, B, and C in the equation of a plane represent the direction of the plane's normal vector. This means that the vector (A, B, C) is perpendicular to the plane, and its length represents the slope of the plane in that direction. This information is useful in understanding the orientation of the plane in relation to other geometric objects.

How can I check if a point lies on a plane given its equation?

To determine if a point (x, y, z) lies on a plane with the equation Ax + By + Cz + D = 0, we can simply substitute the values of x, y, and z into the equation. If the resulting expression is equal to 0, then the point lies on the plane. If it is not equal to 0, then the point is not on the plane.

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

Yes, the equation of a plane can also be written in vector form as r · n = r0 · n, where r is a position vector for any point on the plane, n is the normal vector of the plane, and r0 is a known point on the plane. It can also be written in parametric form as r = r0 + su + tv, where r is a position vector for any point on the plane, r0 is a known point on the plane, and u and v are direction vectors of the plane.

What is the relationship between the equation of a plane and its intercepts?

The intercepts of a plane refer to the points where the plane intersects with the x, y, and z axes. The equation of a plane can be used to find the intercepts by setting one of the variables (x, y, or z) to 0 and solving for the other two variables. For example, setting z = 0 in the equation Ax + By + Cz + D = 0 would give us the x and y intercepts of the plane. Conversely, the intercepts can be used to determine the coefficients in the equation of a plane.

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