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Finding cartesian equation of plane from 3 points 
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#1
Jun1209, 11:18 PM

P: 48

1. The problem statement, all variables and given/known data
Find a Cartesian equation of the plane P containing A (2, 0, −3) , B(1, −1, 6) and C(5, 5, 0) , and determine if point D(3, 2, 3) lies on P. 2. Relevant equations vector cross product ax + by + cz = 0 3. The attempt at a solution Take the cross product of AB and AC to get normal vector. AB = i j + 9k AC = 31 + 5j + 3k I used the determinant method at got: AB X AC = 48i + 30j 2k Now as A, B and C lie on P, take a point say A(2, 0, 3) 48(x  2) +30(y) 2(z + 3) = 0 rearranging that gives: 48x + 30y 2z = 90 Then putting in the x, y and z values for D the equation holds. The question I have is that the answer for the plane given is: 24x − 15y + z = 45 Is there a more common method to follow to get this equation rather than the one I got? 


#2
Jun1209, 11:21 PM

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P: 25,251

The answer is the same one you got. It's just been divided by (2). That doesn't change the plane. E.g. the equation 9x=3 has exactly the same solutions as 3x=1, doesn't it?



#3
Jun1309, 04:33 AM

P: 48

Yes, but why divide by 2? My question was more, how would they get that answer instead? If I am calculating a different (but equivalent) answer then how?



#4
Jun1309, 04:43 AM

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P: 26,160

Finding cartesian equation of plane from 3 points
Hi username12345!
(and the minus because it's customary to start with x, so it makes sense for the x coefficient to be positive) 


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