Find an equation of the plane. The plane through the points (2,1,2), (3,-8,6), and (-2,-3,1)

Wotzdorfg 2021-06-03 Answered
Find an equation of the plane.
The plane through the points (2,1,2), (3,-8,6), and (-2,-3,1)
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Expert Answer

au4gsf
Answered 2021-06-04 Author has 95 answers
To find the equation of the plane we need to find its normal vector.
Let us call the given points as: P(2,1,2), Q(3,-8,6), R(-2,-3,1)
Note that PQ and PR are vectors in the plane
Hence, their cross product will give us normal vector to the plane
PQ=(3,8,6)(2,1,2)=(32,81,62)=(1,9,4)
PR=(2,3,1)(2,1,2)=(22,31,12)=(4,4,1)
n=PQ×PR=|ijk194441|=(25,15,40)
Equation of a plane passing through the point (a,b,c) and having normal vector (l,m,n) is
l(xa)+m(yb)+n(zc)=0
We found the normal vector in the previous cell. For a point on the plane, we can choose any of the three given points, I will choose P(2,1,2)
Equation of the plane is
25(x2)15(y1)40(z2)]=0
Divide throughout by 5
5(x2)3(y1)8(z2)=0
5x3y8z+9=0
Result:
5x3y8z+9=0
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