Question

Show that the following points are coplanar.
(i) (0, −1, 0), (2, 1, −1), (1, 1, 1) and (3, 3, 0)
(ii) (0, 4, 3), (−1, −5, −3), (−2, −2, 1) and (1, 1, −1)

Answer 1

(i) The equation of the plane passing through points (0, −1, 0), (2, 1, −1), (1, 1, 1) is given by

$$\begin{gathered} \left|\begin{array}{ccc}x-0& y+1& z-0\\ 2-0& 1+1& -1-0\\ 1-0& 1+1& 1-0\end{array}\right|=0 \\ \Rightarrow \left|\begin{array}{ccc}x& y+1& z\\ 2& 2& -1\\ 1& 2& 1\end{array}\right|=0 \\ \Rightarrow 4x-3\left(y+1\right)+2z=0 \\ \Rightarrow 4x-3y+2z-3=0 \\ \text{Substituting the last point (3, 3, 0) (it means x = 3; y = 3; z = 0) in this plane equation, we get} \\ 4\left(3\right)-3\left(3\right)+2\left(0\right)-3=0 \\ \Rightarrow 12-12=0 \\ \Rightarrow 0=0 \\ \text{So, the plane equation is satisfied by the point (3, 3, 0).} \\ \text{So, the given points are coplanar.} \end{gathered}$$

(ii) The equation of the plane passing through (0, 4, 3), (−1, −5, −3), (−2, −2, 1) is
$$\begin{gathered} \left|\begin{array}{ccc}x-0& y-4& z-3\\ -1-0& -5-4& -3-3\\ -2-0& -2-4& 1-3\end{array}\right|=0 \\ \Rightarrow \left|\begin{array}{ccc}x& y-4& z-3\\ -1& -9& -6\\ -2& -6& -2\end{array}\right|=0 \\ \Rightarrow -18x+10\left(y-4\right)-12\left(z-3\right)=0 \\ \Rightarrow 9x-5\left(y-4\right)+6\left(z-3\right)=0 \\ \Rightarrow 9x-5y+6z+2=0 \\ \text{Substituting the last point (1, 1, -1) (it means x = 1; y = 1; z=-1) in this plane equation, we get} \\ 9\left(1\right)-5\left(1\right)+6\left(-1\right)+2=0 \\ \Rightarrow 4-4=0 \\ \Rightarrow 0=0 \\ \text{So, the plane equation is satisfied by the point (1, 1, -1).} \\ \text{So, the given points are coplanar.} \end{gathered}$$