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Question 1043423: use a determinent to find an equation of the plane passing through the given
points (1,-2,1)(-1,-1,7)(2,-1,3)
Found 2 solutions by Alan3354, rothauserc:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! use a determinent [sic] to find an equation of the plane passing through the given
points (1,-2,1)(-1,-1,7)(2,-1,3)
|x y z 1|
|1 2 1 1|
|-1 -1 7 1| = 0
|2 -1 3 1|
I can't get the rendering to work, but that's the determinant.
---
x*(2*(7 - 3) - 1*(-1+1) + 1*(-3+7)) - y*(...) .... etc
----
Can you finish it?
Answer by rothauserc(4718)  (Show Source):
You can put this solution on YOUR website! We assume, for now, that the plane does not go through the origin, then
the equation for the plane is
:
ax +by + cz = 1
:
the three equations using the given points are
:
x -2y +z = 1
-x -y +7z = 1
2x -y +3z = 1
:
Use Cramer's rule to solve the following
:
|1 -2 1|
|-1 -1 7| = -27
|2 -1 3|
:
|1 -2 1|
|1 -1 7| = -4 for the "a" column
|1 -1 3|
:
|1 1 1|
|-1 1 7| = 10 for the "b" column
|2 1 3|
:
|1 -2 1|
|-1 -1 1| = -3 for the "c" column
|2 -1 1|
:
a = -4/-27 = 4/27
b = 10/-27 = -10/27
c = -3/-27 = 3/27
:
The equation that contains all three points is
:
(4/27)x -(10/27)y +(3/27)z = 1
:
multiply both sides of = by 27
:
****************
4x -10y +3z = 27
****************
:
we check this by substituting for x, y, z for the given three points
:
(1,-2,1)
4(1) -10(-2) +3(1) = 27
27 = 27
:
(-1,-1,7)
4(-1) -10(-1) +3(7) = 27
-4 +10 +21 = 27
27 = 27
:
(2,-1,3)
4(2) -10(-1) +3(3) = 27
8 +10 +9 = 27
27 = 27
:
our answer checks :-)
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