Class 12 Maths Three Dimensional Geometry | Equation of a plane passing through 3 non collinear points |

**Equation of a plane passing through 3 non collinear points **

Let R, S and T be three non collinear points on the plane with position vectors a^{->}, b^{->} and c^{->} respectively.

This is the equation of plane in vector form passing through three noncollinear points.

These planes resemble the pages of the book where the line containing the points R, S and T are present in the binding of the book.

** Note**:-Three points has to be non-collinear else there will be many planes that will contain them.

__Cartesian form__

Let (x_{1}, y_{1}, z_{1}), (x_{2}, y_{2}, z_{2}) and (x_{3}, y_{3}, z_{3}) be the coordinates of the points R, S and T respectively.

Let (x,y,z) be the coordinates of the point P on the plane with position vector r^{->}.

Then

Substituting these values in equation(1) of the vector form and writing in the determinant form:

The above equation is the equation of the plane in Cartesian form passing through three non collinear points (x_{1}, y_{1}, z_{1}), (x_{2}, y_{2}, z_{2}) and (x_{3}, y_{3}, z_{3}).

__Problem:-__

Find the vector equations of the plane passing through 3 points (1,1,-1),(6,4,5) ,(-4,-2,3)?

__Answer:-__

The given points are (1,1,-1),(6,4,5) ,(-4,-2,3).

=(12-10)-(18-20)-(-12+16)

=2+2-4=0.

Since 3 points are collinear points,there will be infinite number of planes passing through the given points.

__Problem:-__

Find the vector equations of the plane passing through 3 points (1,1,0),(1,2,1) ,(2,2,-1)?

__Answer:-__

The given points are (1,1,0),(1,2,1) ,(2,2,-1).

=(-2-2)-(2+2)=-8≠0.

Therefore, a plane will pass through the points A,B and C.

It is known that the equation of the plane through the points (x_{1}, y_{1}, z_{1}), (x_{2}, y_{2}, z_{2}) and (x_{3}, y_{3}, z_{3}) is

=>(-2) (x-1)-3(y-1) +3z=0

=>-2x-3y+3z+2+3=0

=>2x-3y+3z=-5

=>2x+3y-3z=5.

This is the Cartesian equation of the required plane.

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