Question

Find the vector equation of the plane passing through the points (1, 1, 1), (1, −1, 1) and (−7, −3, −5).

Answer 1

$$\begin{gathered} \text{Let}\mathit{\text{A}}\text{(1, 1, 1),}\mathit{\text{B}}\text{(1, -1, 1) and}\mathit{\text{C}}\text{(-7, -3, -5) be the coordinates.} \\ \text{The required plane passes through the point}\mathit{\text{A}}\text{(1, 1, 1) whose position vector is}\overrightarrow{a}\text{=}\hat{i}\text{+}\hat{j}\text{+}\hat{k}\text{and is normal to the vector}\overrightarrow{n}\text{given by} \\ \overrightarrow{n}=\overrightarrow{AB}\times \overrightarrow{AC.} \\ \text{Clearly,}\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=\left(\hat{i}-\hat{j}+\hat{k}\right)-\left(\hat{i}+\hat{j}+\hat{k}\right)=0\hat{i}-2\hat{j}+0\hat{k} \\ \overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA}=\left(-7\hat{i}-3\hat{j}-5\hat{k}\right)-\left(\hat{i}+\hat{j}+\hat{k}\right)=-8\hat{i}-4\hat{j}-6\hat{k} \\ \overrightarrow{n}=\overrightarrow{AB}\times \overrightarrow{AC}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 0& -2& 0\\ -8& -4& -6\end{array}\right|=12\hat{i}+0\hat{j}-16\hat{k} \\ \text{The vector equation of the required plane is} \\ \overrightarrow{r}.\overrightarrow{n}=\overrightarrow{a}.\overrightarrow{n} \\ \Rightarrow \overrightarrow{r}.\left(12\hat{i}+0\hat{j}-16\hat{k}\right)=\left(\hat{i}\text{+}\hat{j}\text{+}\hat{k}\right).\left(12\hat{i}+0\hat{j}-16\hat{k}\right) \\ \Rightarrow \overrightarrow{r}.\left[4\left(3\hat{i}-4\hat{k}\right)\right]=12+0-16 \\ \Rightarrow \overrightarrow{r}.\left[4\left(3\hat{i}-4\hat{k}\right)\right]=-4 \\ \Rightarrow \overrightarrow{r}.\left(3\hat{i}-4\hat{k}\right)=-1 \\ \Rightarrow \overrightarrow{r}.\left(3\hat{i}-4\hat{k}\right)+1=0 \end{gathered}$$