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 (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 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 (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3).

Problem:-

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

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)?

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 (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3) 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.

.