Equation of plane given points

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To find the equation of a plane containing six given points, the process involves determining two vectors from three of the points and calculating their cross product to obtain the normal vector. The normal vector's coordinates are then used in the plane equation format ax + by + cz = d, where d is derived from substituting one of the points. However, discrepancies arise when testing the equation with additional points, leading to confusion about whether all points lie on the same plane. The discussion highlights potential algebraic errors and questions the validity of the problem, given the number of points provided. Ultimately, the consensus is that the wording of the problem suggests all points should be coplanar, despite the inconsistencies observed.
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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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Check the z-component of your cross product result.
 
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?
 
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.
 
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
 
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?
 
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
 

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