Question

Show that the four points $\left( {0, - 1, - 1} \right),\,\left( {4,5,1} \right),\left( {3,9,4} \right)\,\,and\,\left( { - 4,4,4} \right)$are coplanar.

Answer 1

Hint: To show that given four points are coplanar we first form an equation of plane using three points and then substituting the fourth left point in the equation of a plane formed using three points if fourth points satisfy equation of plane then we can say that all four points are coplanar.
Formulas used: Cartesian equation of a plane $a\left( {x - {x_1}} \right) + b\left( {y - {y_1}} \right) + c\left( {z - {z_1}} \right) = 0$

Complete step-by-step answer:
Given four points are$\left( {0, - 1, - 1} \right),\,\left( {4,5,1} \right),\left( {3,9,4} \right)\,\,and\,\left( { - 4,4,4} \right)$.
We will first form an equation of a plane by using any three points and then substituting the fourth left point in the equation so formed. If points satisfy the plane then all four points are coplanar and if not then they are not coplanar.
Let equation of a plane given as:
$a\left( {x - {x_1}} \right) + b\left( {y - {y_1}} \right) + c\left( {z - {z_1}} \right) = 0$
Where, $\left( {a,b,c} \right)$are direction ratio of the plane and $\left( {{x_1},{y_1},{z_1}} \right)$is a point through which plane passes.
Since, above plane passes through points$\left( {0, - 1, - 1} \right),\,\left( {4,5,1} \right)\,and\,\,\left( {3,9,4} \right)$.
Using all these points in equation of plane we have:
$
  a\left( {x - 0} \right) + b\left( {y + 1} \right) + c\left( {z + 1} \right) = 0 \\
  Also, \\
  a\left( {4 - 0} \right) + b\left( {5 + 1} \right) + c\left( {1 + 1} \right) = 0 \\
  or \\
  4a + 6b + 2c = 0..................(i) \\
  Aslo\,for\,point \,\left( {3,9,4} \right)\,We\,\,have: \\
  a\left( {3 - 0} \right) + b\left( {9 + 1} \right) + c\left( {4 + 1} \right) = 0 \\
  or \\
  3a + 10b + 5c = 0...................(ii) \;
 $
Solving equation (i) and (ii) by cross multiplication method we have.
$
  \dfrac{a}{{\begin{array}{*{20}{l}}
  6&2 \\
  {10}&5
\end{array}}} = \dfrac{b}{{\begin{array}{*{20}{l}}
  2&4 \\
  5&3
\end{array}}} = \dfrac{c}{{\begin{array}{*{20}{l}}
  4&6 \\
  3&{10}
\end{array}}} \\
   \Rightarrow \dfrac{a}{{30 - 20}} = \dfrac{b}{{6 - 20}} = \dfrac{c}{{40 - 18}} \\
   \Rightarrow \dfrac{a}{{10}} = \dfrac{b}{{ - 14}} = \dfrac{c}{{22}} \;
  or \\
  \dfrac{a}{5} = \dfrac{b}{{ - 7}} = \dfrac{c}{{11}} \\
 $
Or we can write
$a = 5k,\,\,b = - 7k\,and\,\,c = 11k$
Substituting values of a, b and c in above equation we have:
$
  5k\left( {x - 0} \right) - 7k\left( {y + 1} \right) + 11k\left( {z + 1} \right) = 0 \\
  or \\
  5\left( x \right) - 7\left( {y + 1} \right) + 11\left( {z + 1} \right) = 0 \\
   \Rightarrow 5x - 7y - 7 + 11z + 11 = 0 \\
   \Rightarrow 5x - 7y + 11z + 4 = 0 \;
 $
Hence, required equation of plane through three points given as: $5x - 7y + 11z + 4 = 0$
Now, substituting the fourth point $\left( { - 4,4,4} \right)$in the above formed equation of plane.
$
  5\left( { - 4} \right) - 7\left( 4 \right) + 11\left( 4 \right) + 4 \\
   \Rightarrow - 20 - 28 + 44 + 4 \\
   \Rightarrow - 48 + 48 \\
   = 0 \;
 $
Therefore, from above we see that the fourth point satisfies the equation of the plane formed by three points.
Hence, we can say that all four points are coplanar.

Note: To discuss given four points are coplanar or not. We can also use the matrix method. In this we first form a matrix using given four points and then if on solving it or simplifying it we get zero then we can say that four points are coplanar.