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Find the equation of a plane given points...?
Find the equation of the plane through the origin and the points (1, −2, 7) and (8, 1, 3).
Okay I've done this twice and my answer is 29(x-1)+53(y+2)+17(z-7)=10
I'm not sure what I'm doing wrong since I have another question after it that is the same thing that is also wrong. I created two vectors using the origin and the other two points, used the cross product to find the normal vector, and then inserted it into the equation for a plane. Am I doing something wrong?
2 Answers
- ?Lv 77 years agoFavorite Answer
The method is described clearly in the link. Following the method gives:
Origin: O (0,0,0)
Point A: (1, -2, 7)
Point B: (8, 1, 3)
OA = 1i -2j + 7k
OB = 8i +1j + 3k7
The normal to the plane is OAXOB:
OAXOB =
|i . j. k|
|1 -2 7|
|8 1 3|
= -13i +53j + 17k
This means the equation of the plane is
-13x + 53y + 17z + d = 0
Since the plane passes through (,0,0,0):
-13*0 + 53*0 + 17*0 + d= 0
d = 0
The final equation is
-13x + 53y + 17z = 0
If you substitute the coordiantes of O, or A or B into this equation, you will find they check.
- alexLv 77 years ago
OA = <1,-2,7>
OB=<8,1,3>
---> normal vector N = OA x OB = <-13 , 53,17>
---> equation of the plane is -13x+53y+17z=0
