Question

Find the equation of the planes that passes through the sets of three points.

(1,1,-1),(6,4,-5) and (-4,-2,3)

(1,1,0),(1,2,1),(-2,2,-1)

Answer 1

The given points are A(1,1,-1), B(6,4,-5) and C(-4,-2,3).

Before determine the equation of the plane , firstly check the condition of collinear.

i.e., $\left|\begin{array}{ccc}{x}_1& y_1& z_1\\ x_2& y_2& z_2\\ x_3& y_2& z_3\end{array}\right|=\left|\begin{array}{ccc}1& 1& -1\\ 6& 4& -5\\ -4& -2& 3\end{array}\right|=1(12-10)$-1(18-20)-1(-12+16)=2+2-4=0

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

The given points are A (1,1,0), B(1,2,1) and C (-2,2,-1).

Firstly, check the condition of collinear point.

$i.e.,\left|\begin{array}{ccc}{x}_1& y_1& z_1\\ x_2& y_2& z_2\\ x_3& y_2& z_3\end{array}\right|=\left|\begin{array}{ccc}1& 1& 0\\ 1& 2& 1\\ -2& 2& -1\end{array}\right|$

=1(-2-2)-1(-1+2)+0(2+4)=-5 $\ne$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

$\left|\begin{array}{ccc}x-x_1& y-y_1& z-z_1\\ x_2-x_1& y_2-y_1& z_2-z_1\\ x_3-x_2& y_3-y_2& z_3-z_2\end{array}\right|=0\Rightarrow \left|\begin{array}{ccc}x-1& y-1& z\\ 0& 1& 1\\ -3& 0& -2\end{array}\right|=0\Rightarrow -2(x-1)-3(y-1)+3z=0\Rightarrow -2x+2-3y+3+3z=0\Rightarrow 2x+3y-3z-5=0\Rightarrow 2x+3y-3z=5$
This is the Cartesian equation of the required plane.