Answer to Question #163677 in Linear Algebra for Nikhil Rawat

Question

Find the vector equation of the plane determined by the points (1,-2,1),(1,0,1) and (1,-1,1). Also check whether (1/2,1/2,1/2) lies on it.

Answer 1

Solution.

If the coordinates of the three points $A(x_1;y_1;z_1),$ $B(x_2;y_2;z_2)$ and $C(x_3;y_3;z_3)$ , which lie on the plane, the plane equation can be found by the following formula

$\begin{vmatrix} x-x_1 & y-y_1 & z-z_1 \\ x_2-x_1 & y_2-y_1 & z_2-z_1 \\ x_3-x_1 & y_3-y_1 & z_3-z_1 \\ \end{vmatrix}=0.$

We will get

$\begin{vmatrix} x-1 & y+2 & z-1 \\ 0 & 2 & 0 \\ 0 & 1 & 0 \\ \end{vmatrix}=0,$

from here

$(x-1)•0+(y+2)•0+(z-1)•0=0, \newline\text{or}\newline 0\cdot x+0\cdot y+0\cdot z+0=0.$

$\vec{n}=\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$ - normal vector.

The vector equation of the plane:

$<\vec{n},\vec{r}>=a,$

where $\vec{n}=\{A;B;C\}=\{0;0;0\},$ $\vec{r}=\{x,y,z\},$ and $a=-D=0.$

So,

$\begin{pmatrix} 0 & 0 & 0 \end{pmatrix} ×\begin{pmatrix} x\\ y\\ z \end{pmatrix}=0$

is the vector equation of the plane. It is not defined. The point (1/2;1/2;1/2) lies on it.