Answer to Question #144752 in Analytic Geometry for Dhruv Rawat

Question

Show that the points (2,0,1), (0,4,-3), (-2,5,0) are non-collinear. Hence the equation of plane passing through them.

Answer 1

$\textsf{Let}\,\, A = (2, 0, 1), B = (0, 4, 3), C = (-2, 5, 0)\\ \textsf{If}\, A.(BC) = 0,\, \textsf{the three points are collinear.} \\ \begin{aligned} \begin{vmatrix} 2 & 0 & 1 \\ 0 & 4 & 3 \\ -2 & 5 & 0 \end{vmatrix} &= 2 \begin{vmatrix} 4 & 3 \\ 5 & 0 \end{vmatrix} + \begin{vmatrix} 0 & 4 \\ -2 & 5 \end{vmatrix} \\&= 2(-15) + 8 = -30 + 8 = -22 \end{aligned} \\ \textsf{Since}\,\, A.(B\times C) \neq 0, \textsf{the three points are}\\ \textsf{not collinear.}\\ \textsf{Also, since we are not given a normal}\\ \textsf{vector, we need to find one by taking}\\ \textsf{the cross product of the displacement vector}\\ \overrightarrow{u}\,\textsf{from}\,\, A \,\textsf{to}\, B \,\, \textsf{and the}\\ \textsf{displacement vector}\,\,\overrightarrow{v} \textsf{from}\,\,B \, \textsf{to} C. \\ \textsf{Then we obtain a vector}\,\overrightarrow{n} \,\textsf{which is}\\ \textsf{normal (orthogonal) to each of the original}\\ \textsf{vectors (and thus orthogonal to the plane).}\\ AB = (0, 4, 3) - (2, 0, 1) = (-2, 4, 2 ) \\ BC = (-2, 5, 0) - (0, 4, 3) = (-2, 1, -3)\\ \overrightarrow{u} \times \overrightarrow{v} = (-2, 4, 2 ) \times (-2, 1, -3)\\ \begin{aligned} \overrightarrow{u} \times \overrightarrow{v} &= \begin{vmatrix} \textbf{i} & \textbf{j} & \textbf{k}\\ -2 & 4 & 2\\ -2 & 1 & -3 \end{vmatrix} \\&= \textbf{i}(-12 - 2) - \textbf{j}(6 + 4) + \textbf{k}(-2 + 8) \\&= -14\textbf{i} - 10\textbf{j} + 6\textbf{k} \end{aligned}\\ \overrightarrow{n} = (-14, -10, 6),\,\, a = (2, 0, 1)\\ \textsf{Hence the equation of the plane is}\\ (\overrightarrow{r} - a)\overrightarrow{n} = 0\\ -14(x - 2) - 10(y - 0) + 6(z - 1) = 0\\ -14x - 10y + 6z = -22$