Maharashtra State BoardHSC Arts 12th Board Exam
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Plane - Equation of a Plane Passing Through Three Non Collinear Points

notes

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

The vectors `vec (RS)` and `vec (RT)` are in the given plane. Therefore , the vector `vec (RS) xx vec (RT)` is perpendicular to the plane containing points R,S and T. Let `vec r`be the position vector of any point P in the plane.  Therefore, the equation of the plane passing through R and  perpendicular to the vector  `vec (RS) xx vec (RT)` is 
`(vec r - vec a) . (vec (RS) xx vec (RT)) = 0`
or `(vec r - vec a) . [(vec b - vec a) xx (vec c - vec 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 `(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 any point P on the plane with position vector `vec r`. Then  
`vec (RP) = (x - x_1) hat i + ( y - y_1) hat j + (z - z_1) hat k`
`vec (RS) = (x_2 - x_1) hat i + ( y_2 - y_1) hat j + (z_2 - z_1) hat k`
`vec (RT) = (x_3 - x_1) hat i + ( y_3 - y_1) hat j + (z_3 - z_1) hat k`
Substituting these values in equation (1) of the vector form and expressing it in the form of a determinant, we have 
`|(x - x_1 , y - y_1 , z - z_1),(x_2 - x_1 , y_2 - y_1, z_2 -z_1) ,(x_3 - x_1, y_3 - y_1, z_3 - z_1) | = 0`
which 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).`

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

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