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Finding cartesian equation of plane from 3 points

by username12345
Tags: cartesian, equation, plane, points
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Jun12-09, 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?
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Jun12-09, 11:21 PM
Sci Advisor
HW Helper
P: 25,235
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?
Jun13-09, 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?

Jun13-09, 04:43 AM
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tiny-tim's Avatar
P: 26,148
Finding cartesian equation of plane from 3 points

Hi username12345!
Quote Quote by username12345 View Post
Yes, but why divide by -2?
Same reason as why we write 3/2 and not 6/4 or 21/14.

(and the minus because it's customary to start with x, so it makes sense for the x coefficient to be positive)
Jun13-09, 05:07 AM
PF Gold
drizzle's Avatar
P: 526
in other words, you do it in the simplest form, about the minus sign,donít you think [guys] itís because the many parts in the equation being negative, so we multiply it by the sign [-]?!

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