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Finding a plane given 3 points

  • Thread starter the7joker7
  • Start date
113
0
1. Homework Statement

Find a plane through the points (6,-7,2), (-4,7,-4), (4,8,-2)

3. The Attempt at a Solution

A = <-10, 14, -6>

B = <-2, 15, -4>

i j k
-10 14 -6
-2 15 -4

i(-146) - j(28) + k(-178)

-146(x - 6) - 28(y + 7) - 178(z + 2)

-146x - 28y - 178z = 324

Actual answer: `34*x -28*y -122*z = 156`

ugh
 

Answers and Replies

73
0
u evaluated the determinant of the cross product matrix incorrectly, it should be 34i-28y-122z, did u use cramer's rule?
 
113
0
Okay, here's my work for a similar problem with the same concept. Does this look right?

P = (-8, -3, 7)

Q = (-8, 7, 4)

R = (-2, -4, -8)

PQ = (0, 10, -3)

PR = (6, -1, -15)

i j k
0 10 -3
6 -1 -15

i(-150 + 3) - j(-18) + k(60) = 0

147(x + 8) + 18(y - 3) + 60(z - 7) = 0

147x + 18y + 60z = -1176 + 54 + 420 = -702
 
113
0
Ugh!!! It's wrong...what am I doing wrong?!
 
73
0
You have the right idea, but perhaps you are calculating the determinant too fast, it should go like this:
i((10*-15)-(-3*-1))-j((0*-15)-(-3*6))+k((0*-1)-(10*6)=
i(-150-3)-j(0-(-18))+k(0-60)=
-153i-18j-60k
Now, as you are having the right idea, take the dot product of our newfound normal vector with P and set it equal to zero since, by definition, it is normal:
(-153i-18j-60k).(i(x-(-8))+j(y-(-3))+k(z-7))=
153(x+8)-18(y+3)-60(z-7)=0
You should be able to solve it from there. Let me know if this answer does not coincide with the book. There are lots of double negatives that you seemed to have misplaced in your calculations...be careful!
 

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