Question

Show that the points \[\left( -1,4,-3 \right),\left( 3,2,-5 \right),\left( -3,8,-5\text{ } \right)\text{and }\left( -3,2,1 \right)\] are coplanar?

Answer 1

Hint: In this question, we have to prove that the given points are coplanar. Thus, we will use the vector-coplanar formula to get the solution. As we know, coplanar means when the lines or the planes lie on the same plane. Thus, to prove this problem, we have to show that the value of determinant is equal to 0. Therefore, we start solving this problem, by converting the points into vector form and then taking three vectors AB, BC, and CD. Then, we will find the value of these vectors. After that, we will find the determinant of these vectors using the determinant formula, to get the required solution for the problem.

Complete step by step solution:
According to the problem, we have to prove that the given points are coplanar.
Thus, we will use the vector-coplanar formula to get the solution.
The points give to us is \[\left( -1,4,-3 \right),\left( 3,2,-5 \right),\left( -3,8,-5\text{ } \right)\text{and }\left( -3,2,1 \right)\] ------- (1)
Now, we will first convert the points from equation (1) into the vector form, that is thwe get
$\overrightarrow{OA}=-1\hat{i}+4\hat{j}-3\hat{k}$ --------- (2)
$\overrightarrow{OB}=3\hat{i}+2\hat{j}-5\hat{k}$ --------- (3)
$\overrightarrow{OC}=-3\hat{i}+8\hat{j}-5\hat{k}$ --------- (4)
$\overrightarrow{OD}=-3\hat{i}+2\hat{j}+1\hat{k}$ ---------- (5)
Now, the three vectors from A, B, C, and D are $\overrightarrow{AB},\overrightarrow{BC},\text{ and }\overrightarrow{CD}$
Thus, we will now find the values of these three vectors, we get
$\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}$
Now, we will put the value of equation (2) and (3) in the above equation, we get
$\Rightarrow \overrightarrow{AB}=3\hat{i}+2\hat{j}-5\hat{k}-\left( -1\hat{i}+4\hat{j}-3\hat{k} \right)$
On further solving the above equation, we get
$\Rightarrow \overrightarrow{AB}=3\hat{i}+2\hat{j}-5\hat{k}+1\hat{i}-4\hat{j}+3\hat{k}$
$\Rightarrow \overrightarrow{AB}=4\hat{i}-2\hat{j}-2\hat{k}$ ---------- (6)
Now, the value of another vector, we get
$\overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA}$
Now, we will put the value of equation (2) and (4) in the above equation, we get
$\Rightarrow \overrightarrow{AC}=-3\hat{i}+8\hat{j}-5\hat{k}-\left( -\hat{i}+4\hat{j}-3\hat{k} \right)$
On further solving the above equation, we get
$\Rightarrow \overrightarrow{AC}=-3\hat{i}+8\hat{j}-5\hat{k}+\hat{i}-4\hat{j}+3\hat{k}$
$\Rightarrow \overrightarrow{AC}=-2\hat{i}+4\hat{j}-2\hat{k}$ ---------- (7)
Now, the value of another vector, we get
$\overrightarrow{AD}=\overrightarrow{OD}-\overrightarrow{OA}$
Now, we will put the value of equation (2) and (5) in the above equation, we get
$\Rightarrow \overrightarrow{AD}=-3\hat{i}+2\hat{j}+1\hat{k}-\left( -1\hat{i}+4\hat{j}-3\hat{k} \right)$
On further solving the above equation, we get
$\Rightarrow \overrightarrow{AD}=-3\hat{i}+2\hat{j}+1\hat{k}+\hat{i}-4\hat{j}+3\hat{k}$
$\Rightarrow \overrightarrow{AD}=-2\hat{i}-2\hat{j}+4\hat{k}$ ---------- (8)
Now, we will find the determinant of these three vectors from equation (6), (7), and (8), we get
$\left[ \overrightarrow{AB},\overrightarrow{BC},\overrightarrow{CD} \right]=\left[ \begin{matrix}
   4 & -2 & -2 \\
   -2 & 4 & -2 \\
   -2 & -2 & 4 \\
\end{matrix} \right]$
So, we will apply the formula of determinant. Let us suppose a matrix A which is equal to
$A=\left[ \begin{matrix}
   a & d & g \\
   b & e & h \\
   c & f & i \\
\end{matrix} \right]$
Then, the value of determinant is equal to
$a\left[ \begin{matrix}
   e & h \\
   f & i \\
\end{matrix} \right]-\left( d \right)\left[ \begin{matrix}
   b & h \\
   c & i \\
\end{matrix} \right]+\left( g \right)\left[ \begin{matrix}
   b & e \\
   c & f \\
\end{matrix} \right]$
Thus, we get
$\Rightarrow 4\left[ \begin{matrix}
   4 & -2 \\
   -2 & 4 \\
\end{matrix} \right]-\left( -2 \right)\left[ \begin{matrix}
   -2 & -2 \\
   -2 & 4 \\
\end{matrix} \right]+\left( -2 \right)\left[ \begin{matrix}
   -2 & 4 \\
   -2 & -2 \\
\end{matrix} \right]$
Now, we know that the value of determinant in 2X2 matrix is equal to the subtraction of the diagonals, such that the first diagonal is the multiplication of the elements from top left to below right and second diagonal is equal to the multiplication of element from top right to below left, thus the above equation will equal to
$\Rightarrow 4\left[ 4.\left( 4 \right)-\left( -2 \right).\left( -2 \right) \right]-\left( -2 \right)\left[ -2.\left( 4 \right)-\left( -2 \right)\left( -2 \right) \right]+\left( -2 \right)\left[ \left( -2 \right).\left( -2 \right)-\left( 4 \right).\left( -2 \right) \right]$
On solving the brackets of the above equation, we get
$\Rightarrow 4\left[ 16-4 \right]-\left( -2 \right)\left[ -8-4 \right]+\left( -2 \right)\left[ 4+8 \right]$
$\Rightarrow 4\left[ 12 \right]-\left( -2 \right)\left[ -12 \right]+\left( -2 \right)\left[ 12 \right]$
On opening the brackets of the above equation, we get
$\Rightarrow 48-24-24$
$\Rightarrow 48-48$
$\Rightarrow 0$
Thus, the value of determinant is equal to 0, therefore the vectors are coplanar, that is they lie on the same plane.
Therefore, the points \[\left( -1,4,-3 \right),\left( 3,2,-5 \right),\left( -3,8,-5\text{ } \right)\text{and }\left( -3,2,1 \right)\] are coplanar.

Note: While solving this problem, do mention all the steps properly to avoid mathematical error and confusion. Do not forget to mention the definition of the coplanar. Always remember that in determinant, the value is equal to zero if any two rows and columns are equal to each other.