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Cartesian Equation of a plane containing 3 points

  • Thread starter Blackh4wk
  • Start date
  • #1
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Homework Statement


Find the Cartesian Eqationn of the plane containing the points (0,-2,6), (1,8,-2) and (2,0,5).



Homework Equations





The Attempt at a Solution



OA = 0,-2,6


AB = 1,10,8


AC = 2,2,-1

Cross product of AB AC = 6,-15,-18

X - 0 * 6
Y - (-2) * -15
Z - 6 * -18

Which becomes,

6x - 15y + 30 - 18z + 108 = 0

Clean it up

6x - 15y - 18z = -138

Proper answer is

2x - 5y - 6z = -26

Now, I know how you get the LHS to = the proper answer (divide by 3), but how do I go about getting the RHS to equate properly?

Or have I screwed up some of my arithmetic?
 

Answers and Replies

  • #2
SammyS
Staff Emeritus
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Homework Helper
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Homework Statement


Find the Cartesian Eqationn of the plane containing the points (0,-2,6), (1,8,-2) and (2,0,5).

Homework Equations



The Attempt at a Solution



OA = 0,-2,6


AB = 1,10,8


AC = 2,2,-1

Cross product of AB AC = 6,-15,-18

X - 0 * 6
Y - (-2) * -15
Z - 6 * -18

Which becomes,

6x - 15y + 30 - 18z + 108 = 0

Clean it up

6x - 15y - 18z = -138

Proper answer is

2x - 5y - 6z = -26

Now, I know how you get the LHS to = the proper answer (divide by 3), but how do I go about getting the RHS to equate properly?

Or have I screwed up some of my arithmetic?
Hello Blackh4wk. Welcome to PF !

No, your arithmetic looks OK, but you found the constant incorrectly.

Once you find [itex]\displaystyle \stackrel{\rightarrow}{\text{AB}}\times\stackrel{ \rightarrow}{\text{AC}}=\left\langle 6,\,-15,\,-18 \right\rangle\,,[/itex] and notice that each of the components is a multiple of 3, you should know that the equation of the plane can be written as:
[itex]\displaystyle 2x-5y-6z=d\,,[/itex]​
where d is a constant. Plugging in the coordinates of any point in the plane should give you d.
 

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