Question

Show that the four points $A(4,5,1),B(0,-1,-1),C(3,9,4)\text{ and }D(-4,4,4)$ are coplanar.

Answer 1

Hint: In the given question we have to prove that the four given points $A(4,5,1),B(0,-1,-1),C(3,9,4)\text{ and }D(-4,4,4)$ are coplanar. That is all the given points lies on the same plane. As we know that for a plane to be find we need at least three points. So, first of all we have to find the equation of plane passes through any three$A(4,5,1),B(0,-1,-1),C(3,9,4)$of the given four points. After that we have to show that the fourth point $\text{ }D(-4,4,4)$satisfy the equation obtained the plane. The general equation of plane passing through a point $({{x}_{1}},{{y}_{1}},{{z}_{1}})$ is
$a(x-{{x}_{1}})+b(y-{{y}_{1}})+c(z-{{z}_{1}})=0$, where $a,b,c$are constants, which is to be find. Form three equation by substituting the three points given, then find the value of $a,b,c$ in order to obtain the equation of the plane.

Complete step by step answer:
The general equation of a plane passing through a point $A(4,5,1)$ is given by
$a(x-4)+b(y-5)+c(z-1)=0-----(1)$
Since it passes through the point $B(0,-1,-1)\text{ and }C(3,9,4)$ we have
\[\begin{align}
  & a(0-4)+b(-1-5)+c(-1-1)=0 \\
 & -4a-6b-2c=0 \\
 & 2a+3b+c=0-------(2) \\
\end{align}\]
Also,
\[\begin{align}
  & a(3-4)+b(9-5)+c(4-1)=0 \\
 & -a+4b+3c=0-------(3) \\
\end{align}\]
Now we have to solve the equation $(2)\text{ and (3)}$ by using cross multiplication method
So, we have
\[\begin{align}
  & \dfrac{a}{(3)(3)-(1)(4)}=\dfrac{-b}{(2)(3)-(1)(-1)}=\dfrac{c}{(2)(4)-(-1)(3)} \\
 & \Rightarrow \dfrac{a}{5}=\dfrac{-b}{7}=\dfrac{c}{11} \\
 & \Rightarrow \dfrac{a}{5}=\dfrac{-b}{+7}=\dfrac{c}{11}=k(\text{say)} \\
\end{align}\]
So, we can write
$\begin{align}
  & a=5k \\
 & b=-7k \\
 & c=11k \\
\end{align}$
Now we have to substitute the value of $a,b,c$ in equation $(1)$we can write
$\begin{align}
  & 5k(x-4)-7k(y-5)+11k(z-1)=0 \\
 & 5(x-4)-7(y-5)+11(z-1)=0 \\
 & \Rightarrow 5x-20-7y+35+11z-11=0 \\
 & \Rightarrow 5x-7y+11z=-4 \\
\end{align}$
Now transposing constant terms to right hand side, we can write
$5x-7y+11z=-4$
So, we get the equation of plane.
Now if the point $\text{ }D(-4,4,4)$satisfy the equation of plane so obtained, the point D also lies on the plane.
Substituting $\text{ }D(-4,4,4)$in equation of plane
\[LHS=5(-4)-7\left( 4 \right)+11(4)\]
$LSH=-4=RHS$
As we see here the point D satisfy the equation of the plane so all the given four points are coplanar.

Note:
This question can also be solved by using vector analysis. For this form vector $\overrightarrow{AB},\overrightarrow{AC}\text{ and }\overrightarrow{\text{AD}}$.
We can write $\begin{align}
  & \overrightarrow{AB}=-4\overrightarrow{i}-6\overrightarrow{j}-2\overrightarrow{k} \\
 & \overrightarrow{AC}=-\overrightarrow{i}+4\overrightarrow{j}+3\overrightarrow{k} \\
 & \overrightarrow{AD}=-8\overrightarrow{i}+\overrightarrow{j}+3\overrightarrow{k} \\
\end{align}$
Geometrically these vectors form a three-dimensional figure. As the given vectors are coplanar its volume is zero.
So, calculate $\left| \overrightarrow{AB}\text{ }\overrightarrow{\text{AC}}\text{ }\overrightarrow{\text{AD}} \right|=\left| \begin{matrix}
   -4 & -6 & -2 \\
   -1 & 4 & 3 \\
   -8 & 1 & 3 \\
\end{matrix} \right|$
If the above determinant is zero it means the given points are coplanar.