# How do we find out whether four points A(3,-1,-1),B(-2,1,2), C(8,-3,0) and D(0,2,-1) lie in the same plane or not?

##### 1 Answer
Feb 24, 2018

$A , B , C$ and $D$ do not lie in the same plane.

#### Explanation:

Three non-collinear points are always define a plane. If fourth plane too is on this plane, four plane define this plane. So let us first define a plane using points $A \left(3 , - 1 , - 1\right) , B \left(- 2 , 1 , 2\right)$ and $D \left(0 , 2 , - 1\right)$, using $\vec{A B} = \left({B}_{x} - {A}_{x}\right) \hat{i} + \left({B}_{y} - {A}_{y}\right) \hat{j} + \left({B}_{z} - {A}_{z}\right) \hat{k}$

Therefore $\vec{A B} = \left(- 2 - 3\right) \hat{i} + \left(1 - \left(- 1\right)\right) \hat{j} + \left(2 - \left(- 1\right)\right) \hat{k} =$

= $- 5 \hat{i} + 2 \hat{j} + 3 \hat{k}$ and

$\vec{A D} = \left(0 - 3\right) \hat{i} + \left(2 - \left(- 1\right)\right) \hat{j} + \left(- 1 - \left(- 1\right)\right) \hat{k} =$

= $- 3 \hat{i} + 3 \hat{j} + 0 \hat{k}$

If $\vec{A B}$ and $\vec{A D}$ are in the same plane, then we will have $\vec{A B} \times \vec{A D} = 0$, the cross product of the two vector as $0$ and hence

$| \left(\hat{i} , \hat{j} , \hat{k}\right) , \left(- 5 , 2 , 3\right) , \left(- 3 , 3 , 0\right) | = 0$

or $\left(0 - 9\right) \hat{i} - \left(0 - \left(- 9\right)\right) \hat{j} + \left(- 15 - \left(- 6\right)\right) \hat{k} = 0$

or $- 9 \hat{i} - 9 \hat{j} - 9 \hat{k} = 0$

or $\hat{i} + \hat{j} + \hat{k} = 0$

Hence equation of plane is $x + y + z = k$ and putting values of points $A , B$ and $D$, we get $k = 1$

Hence equation of plane is $x + y + z = 1$

and as $C \left(8 , - 3 , 0\right)$ does not satisfy it,

$A , B , C$ and $D$ do not lie in the same plane.