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Equation of a Plane - Equation of a Plane Passing Through Three Non Collinear Points

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Notes

Let R, S and T be three non collinear points on the plane with position vectors a , b and c respectively in following fig.

The vectors RS and RT are in the given plane. Therefore , the vector RS×RT is perpendicular to the plane containing points R,S and T. Let rbe the position vector of any point P in the plane.  Therefore, the equation of the plane passing through R and  perpendicular to the vector  RS×RT is 
(r-a).(RS×RT)=0
or (r-a).[(b-a)×(c-a)]=0  ...(1)
If the three points were on the same line, then there will be many planes that will contain them Fig.

for example , These planes will resemble the pages of a book where the line containing the points R, S and T are members in the binding of the book.

Cartesian form:
Let (x1,y1,z1),(x2,y2,z2) and (x3,y3,z3)  be the coordinates of the points R, S and T respectively.  Let (x, y, z) be the coordinates of any point P on the plane with position vector r. Then  
RP=(x-x1)i^+(y-y1)j^+(z-z1)k^
RS=(x2-x1)i^+(y2-y1)j^+(z2-z1)k^
RT=(x3-x1)i^+(y3-y1)j^+(z3-z1)k^
Substituting these values in equation (1) of the vector form and expressing it in the form of a determinant, we have 
|x-x1y-y1z-z1x2-x1y2-y1z2-z1x3-x1y3-y1z3-z1|=0
which is the equation of the plane in Cartesian form passing through three non collinear points (x1,y1,z1),(x2,y2,z2) and (x3,y3,z3).

Video link : https://youtu.be/PCyo3E5kOcw

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