Answer to Question #104110 in Analytic Geometry for Deepak Rana

Question

Check whether the points (1,−1,−2),(1,−4,2),(3,0,2),(4,−3−2) are coplanar
or not. If they are coplanar, write the equation of the plane they pass through.
Otherwise, change the coordinates of one of the points so that they become
coplanar. In this case, find the plane passing through them.

Answer 1

Let $A(1,-1,-2), B(1,-4,2),C(3,0,2),D(4,-3,-2)$

Create 3 vectors $\overrightarrow{AB}, \overrightarrow{AC},$ and $\overrightarrow{AD}$

$\overrightarrow{AB}=(1-1,-4-(-1),2-(-2))=(0,-3,4)$

$\overrightarrow{AC}=(3-1,0-(-1),2-(-2))=(2,1,4)$

$\overrightarrow{AD}=(4-1,-3-(-1),-2-(-2))=(3,-2,0)$

$$\overrightarrow{AB}\cdot(\overrightarrow{AC}\times \overrightarrow{AD})=\begin{vmatrix} b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \\ d_1 & d_2 & d_3 \end{vmatrix}=$$

$$=\begin{vmatrix} 0 & -3 & 4\\ 2 & 1 & 4 \\ 3 & -2 & 0 \end{vmatrix}=0-2\begin{vmatrix} -3 & 4 \\ -2 & 0 \end{vmatrix}+3\begin{vmatrix} -3 & 4 \\ 1 & 4 \end{vmatrix}=$$

$$=-2(-3(0)-4(-2))+3(-3(4)-4(1))=$$

$$=-16-48=-64\not=0$$

The vectors are not coplanar, as their scalar triple product is not equal to zero.

$$ax+by+cz+d=0$$

$A(1,-1,-2): a-b-2c+d=0$

$C(3,0,2): 3a+2c+d=0$

$D(4,-3,-2): 4a-3b-2c+d=0$

$4a-3b-2c+d-a+b+2c-d=0=>3a=2b$

$3a+2c+d-a+b+2c-d=0=>4c=-2a-b$

Let $a=8.$ Then $b=12, 4c=-16-12=-28=>c=-7$

$8-12-2(-7)+d=0=>d=-10$

The plane passing through three points.

$$8x+12y-7z-10=0$$

Check for $B^*(2,3,6)$

$$8(2)+12(3)-7(6)-10=16+36-42-10=0, True$$

$$8x+12y-7z-10=0$$

$A(1,-1,-2), B^*(2,3,6),C(3,0,2),D(4,-3,-2)$