Question

How do I find the cartesian equation of a plane containing three given points?

Answer 1

Here in this question, we have to find the cartesian form of equation of a plane containing three given points. Let's consider any plane. Let A, B and C be three noncollinear points and define two vectors $\overrightarrow {AB}$ and $\overrightarrow {AC}$ contained in a plane. By the Cross product of these two vectors we get the required cartesian equation of the plane.

Complete step by step answer:
A plane is a flat surface with no thickness. The Cartesian Plane is sometimes referred to as the x-y plane or the coordinate plane and is used to plot data pairs on a two-line graph. The Cartesian plane is named after the mathematician Rene Descartes who originally came up with the concept. Cartesian planes are formed by two perpendicular number lines intersecting.
Let us consider the plane and three non collinear points A, B and C, non collinear points refer to those points that do not all lie on the same line. Let the position vectors of these points be $\overrightarrow a$ , $\overrightarrow b$ , and $\overrightarrow c$. We know that the product of these position vectors are $\overrightarrow {AB}$ and $\overrightarrow {AC}$. must be the cross product of two vectors contained in a plane defines the normal vector of the plane.
Let us now consider the coordinates of the three non collinear points as
$\left( {A\left( {{x_1},{y_1},{z_1}} \right),B\left( {{x_2},{y_2},{z_2}} \right),C\left( {{x_3},{y_3},{z_3}} \right)} \right)$
Let P be a point on the plane that must contain these points. The position vector of the point P be $\overrightarrow r$ and its coordinates are $\left( {x,y,z} \right)$. Now, we can write the vectors as
$\overrightarrow {AP} = \left( {x - {x_1}} \right)\widehat i + \left( {y - {y_1}} \right)\widehat j + \left( {z - {z_1}} \right)\widehat k$
$\overrightarrow {AB} = \left( {{x_2} - {x_1}} \right)\widehat i + \left( {{y_2} - {y_1}} \right)\widehat j + \left( {{z_2} - {z_1}} \right)\widehat k$
And
$\overrightarrow {AC} = \left( {{x_3} - {x_1}} \right)\widehat i + \left( {{y_3} - {y_1}} \right)\widehat j + \left( {{z_3} - {z_1}} \right)\widehat k$
We can now use the Vector form to substitute the above vectors. Doing so would give us the advantage of presenting the equation in determinant form as

{\left( {x - {x_1}} \right)}&{\left( {y - {y_1}} \right)}&{\left( {z - {z_1}} \right)} \\\ {\left( {{x_2} - {x_1}} \right)}&{\left( {{y_2} - {y_1}} \right)}&{\left( {{z_2} - {z_1}} \right)} \\\ {\left( {{x_3} - {x_1}} \right)}&{\left( {{y_3} - {y_1}} \right)}&{\left( {{z_3} - {z_1}} \right)} \end{array}} \right| = 0$$ Thus the determinant form gives us nothing but the Cartesian equation of a plane passing through three non collinear points. **Note:** In the representing the coordinates we have two different forms, one is cartesian form and the other one is polar form. In the cartesian form the coordinates are mentioned in the form of x and y. . In the polar form the coordinates are mentioned in the form of r and theta. This method is a general method.