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

Wotzdorfg

Wotzdorfg

Answered question

2021-06-03

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

Answer & Explanation

au4gsf

au4gsf

Skilled2021-06-04Added 95 answers

We must locate the plane's normal vector in order to determine its equation.
We'll refer to the indicated positions as P(2,1,2), Q(3,-8,6), and R. (-2,-3,1)
Note that PQ and PR are vectors in the plane 
Consequently, their cross product will provide us with the plane's normal vector.
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 through (a,b,c) with normal vector (l,m,n) is
l(xa)+m(yb)+n(zc)=0 
In the preceding cell, we located the normal vector. We can select any one of the three locations on the plane, so I'll go with P. (2,1,2)
The plane's equation 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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