Question

Are the four points \[A\left( {1, - 1,1} \right),B\left( { - 1,1,1} \right),C\left( {1,1,1} \right)\,\,{\text{and}}\,\,D\left( {2, - 3,4} \right)\] are coplanar? Justify your answer.

Answer 1

Hint: In this problem, first we need to find the position vectors of the given points. Next, find the vectors joining the points \[AB,AC\,\,{\text{and}}\,\,AD\]. Then, apply the condition of the points to be coplanar. Four points will be coplanar if the volume generated by the points is zero.

Complete step by step answer:
The given points are \[A\left( {1, - 1,1} \right),B\left( { - 1,1,1} \right),C\left( {1,1,1} \right)\,\,{\text{and}}\,\,D\left( {2, - 3,4} \right)\].
The position vectors of the points \[A,B,C,D\] are shown below.
\[
  \vec a = \hat i - \hat j + \hat k \\
  \vec b = - \hat i + \hat j + \hat k \\
  \vec c = \hat i + \hat j + \hat k \\
  \vec d = 2\hat i - 3\hat j + 4\hat k \\
\]
The vector joining the points \[A\] and \[B\] is calculated as follows:
\[
  \,\,\,\,\,\overrightarrow {AB} = \vec b - \vec a \\
   \Rightarrow \overrightarrow {AB} = - \hat i + \hat j + \hat k - \hat i + \hat j - \hat k \\
   \Rightarrow \overrightarrow {AB} = - 2\hat i + 2\hat j \\
\]
The vector joining the points \[A\] and \[C\] is calculated as follows:
\[
  \,\,\,\,\,\overrightarrow {AC} = \vec c - \vec a \\
   \Rightarrow \overrightarrow {AC} = \hat i + \hat j + \hat k - \hat i + \hat j - \hat k \\
   \Rightarrow \overrightarrow {AC} = 2\hat j \\
\]
The vector joining the points \[A\] and \[D\] is calculated as follows:
\[
  \,\,\,\,\,\overrightarrow {AD} = \vec d - \vec a \\
   \Rightarrow \overrightarrow {AD} = 2\hat i - 3\hat j + 4\hat k - \hat i + \hat j - \hat k \\
   \Rightarrow \overrightarrow {AD} = \hat i - 2\hat j + 3\hat k \\
\]
For, points \[A,B,C,D\] to be coplanar, there exist scalars \[x,y\] such that,
\[
  \,\,\,\,\,\,\overrightarrow {AB} = x \cdot \overrightarrow {AC} + y \cdot \overrightarrow {AD} \\
   \Rightarrow - 2\hat i + 2\hat j = x\left( {2\hat j} \right) + y\left( {\hat i - 2\hat j + 3\hat k} \right) \\
   \Rightarrow - 2\hat i + 2\hat j = y\hat i + \left( {2x - 2y} \right)\hat j + 3y\hat k \\
\]
By equation the vectors,
\[
  y = - 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......\left( 1 \right) \\
  2x - 2y = 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......\left( 2 \right) \\
  3y = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......\left( 3 \right) \\
\]
From equation (1), \[y = - 2\], and from equation (3), \[y = 0\] which is not possible.
Thus, the points \[A,B,C,D\] are not coplanar.

Note: In this problem, we need to form three vectors using the given points. Further, we need to find if there exist any linear relationship or not.