Question

Find the vector equation of the plane passing through the points (1, 1, −1), (6, 4, −5) and (−4, −2, 3).

Answer 1

$$\begin{gathered} \text{Let}\mathit{\text{A}}\text{(1, 1, -1),}\mathit{\text{B}}\text{(6, 4, -5) and}\mathit{\text{C}}\text{(-4, -2, 3).} \\ \text{The required plane passes through the point}\mathit{\text{A}}\text{(1, 1, -1) whose position vector is}\overrightarrow{a}\text{=}\hat{i}\text{+}\hat{j}\text{-}\hat{k}\text{and is normal to the vector}\overrightarrow{n}\text{given by} \\ \overrightarrow{n}=\overrightarrow{AB}\times \overrightarrow{AC} \\ \text{Clearly,}\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=\left(6\hat{i}+4\hat{j}-5\hat{k}\right)-\left(\hat{i}+\hat{j}-\hat{k}\right)=5\hat{i}+3\hat{j}-4\hat{k} \\ \overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA}=\left(-4\hat{i}-2\hat{j}+3\hat{k}\right)-\left(\hat{i}+\hat{j}-\hat{k}\right)=-5\hat{i}-3\hat{j}+4\hat{k} \\ \overrightarrow{n}=\overrightarrow{AB}\times \overrightarrow{AC}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 5& 3& -4\\ -5& -3& 4\end{array}\right|=0\hat{i}+0\hat{j}+0\hat{k}=\overrightarrow{0} \\ \text{So, the given points are collinear.} \\ \text{Thus, there will be infinite number of planes passing through these points.} \\ \text{Their equations (passing through (1, 1, -1) are given by} \\ a\left(x-1\right)+b\left(y-1\right)+c\left(z+1\right)=0...\left(1\right) \\ \text{Since this passes through}\mathit{\text{B}}\text{(6, 4, -5),} \\ a\left(6-1\right)+b\left(4-1\right)+c\left(-5+1\right)=0 \\ \Rightarrow 5a+3b-4c=0...\left(2\right) \\ \text{From (1) and (2), the equations of the infinite planes are} \\ a\left(x-1\right)+b\left(y-1\right)+c\left(z+1\right)=0,\text{where}5a+3b-4c=0. \end{gathered}$$