#### 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