Answer to Question #198762 in Analytic Geometry for Dhruv bartwal

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:-

points  (-1,-2,1),(1,0,1) and (1,-1,1).

we will use this formula

$\begin{vmatrix} x - x_A & y - y_A&z - z_A \\ x_B - x_A & y_B - y_A & z_B - z_A\\ x_C - x_A &y_C - y_A & z_C - z_A \end{vmatrix}=0$

on putting the values

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

$\implies$

$[x - (-1)][2·0-0·1]-[y - (-2)][2·0-0·2]+[z - 1][2·1-2·2]=0$

$\implies$

$\boxed{- 2z + 2 = 0}$

this is the required equation

Now we check the points $(\frac{1}{2},\frac{1}{2},\frac{1}{2})$

$=-2\times\frac{1}{2}+2\\ =-1+2\\ =1 \ne0$

$\implies$ points does not lie on plane.