Equation of plane given points

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Homework Help Overview

The problem involves finding the equation of a plane that contains multiple points in three-dimensional space: (1,1,5), (3,5,3), (8,8,1), (10,2,2), (18,6,-1), and (-1,-3,6). The original poster attempts to use vectors derived from three of these points to determine the normal vector of the plane and subsequently formulate the plane's equation.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss calculating vectors from points and using the cross product to find the normal vector. There are attempts to validate the resulting plane equation with various points, leading to questions about the consistency of the points being coplanar.

Discussion Status

Participants are actively engaging with the problem, checking calculations, and questioning the assumptions about the points being coplanar. There is no explicit consensus on the validity of the plane equation derived, as discrepancies arise when testing different points.

Contextual Notes

There is uncertainty regarding whether all six points are intended to lie on the same plane, given the nature of the problem and the number of points provided.

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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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