Answer to Question #114945 in Analytic Geometry for ANJU JAYACHANDRAN

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)

Then, $\vec{AB}$ = (0,3,-4)

$\vec{AC}$ = (2,1,4)

$\vec{AD}$ = (3,-2,0)

To check whether A,B,C,D are coplanar:

$\vec{AC} × \vec{AB} = \begin{vmatrix} i & j & k \\ 2 & 1 & 4 \\ 0 & -3 & 4 \end{vmatrix}$ = 16i -8j -6k

$\vec{AD} . (\vec{AC} × \vec{AB}) = 3(16)+2(8)+0=64$

Which is not equal to zero, so the points A,B,C,D are not coplanar.

To change the coordinates so that A,B,C,D become coplanar:

Let the equation of the plane be

ax+by+cz+d =0

If the plane passes through point A, then,

a-b-2c+d=0

If the plane passes through point B, then,

a-4b+2c+d=0

If the plane passes through point C, then,

3a+2c+d=0

Solving the above three equations,

We get:

a=b

d=2c

3b=4c

Let d=6

Then, c=3, and a=b= 4

So the coordinates of new point D=(4,4,3)

To find the equation of plane passing through

A(1,-1,-2)

B (1,-4,2)

C (3,0,2)

D ( 4,4,3)

$\vec{DC} = (-1,-4,-1)$

$\vec{DB} = (-3,-8,-1)$

$\begin{vmatrix} x-4 & -1 & -3\\ y-4 & -4 & -8\\ z-3 & -1 & -1 \end{vmatrix}$ =0

4x - 2y +4z = 20

2x -y + 2z = 10