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Question
Show that the four points (0, −1, −1), (4, 5, 1), (3, 9, 4) and (−4, 4, 4) are coplanar and find the equation of the common plane.
Solution
The equation of the plane passing through the points (0, −1, −1), (4, 5, 1) and (3, 9, 4) is given by
\[\begin{vmatrix}x - 0 & y + 1 & z + 1 \\ 4 - 0 & 5 + 1 & 1 + 1 \\ 3 - 0 & 9 + 1 & 4 + 1\end{vmatrix} = 0\]
\[ \Rightarrow \begin{vmatrix}x & y + 1 & z + 1 \\ 4 & 6 & 2 \\ 3 & 10 & 5\end{vmatrix} = 0\]
\[ \Rightarrow 10x - 14 \left( y + 1 \right) + 22 \left( z + 1 \right) = 0\]
\[ \Rightarrow 5x - 7 \left( y + 1 \right) + 11 \left( z + 1 \right) = 0\]
\[ \Rightarrow 5x - 7y + 11z + 4 = 0\]
\[\text{ Substituting the last point } (-4, 4, 4) (\text{ it means } x = -4; y = 4; z = 4) \text{ in this plane equation, we get } \]
\[5 \left( - 4 \right) - 7 \left( 4 \right) + 11 \left( 4 \right) + 4 = 0\]
\[ \Rightarrow - 48 + 48 = 0\]
\[ \Rightarrow 0 = 0\]
\[\text{ So, the plane equation is satisfied by the point } (-4, 4, 4).\]
\[\text{ So, the given points are coplanar and the equation of the common plane (as we already found) is } \]
\[5x - 7y + 11z + 4 = 0 . \]
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