Question

Find the vector equation of the plane passing through the points $3\hat{i}+4\hat{j}+2\hat{k},2\hat{i}-2\hat{j}-\hat{k}\mathrm{and}7\hat{i}+6\hat{k}.$

Answer 1

$$\begin{gathered} \text{Let}\mathit{\text{A}}\text{(3, 4, 2),}\mathit{\text{B}}\text{(2, -2, -1) and}\mathit{\text{C}}\text{(7, 0, 6) be the points represented by the given position vectors.} \\ \text{The required plane passes through the point}\mathit{\text{A}}(3,\; 4,\; 2)\text{whose position vector is}\overrightarrow{a}\text{=3}\hat{i}\text{+4}\hat{j}\text{+2}\hat{k}\text{and is normal to the vector}\overrightarrow{n}\text{given by} \\ \overrightarrow{n}=\overrightarrow{AB}\times \overrightarrow{AC}. \\ \text{Clearly,}\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=\left(2\hat{i}-2\hat{j}-\hat{k}\right)-\left(3\hat{i}\text{+4}\hat{j}\text{+2}\hat{k}\right)=-\hat{i}-6\hat{j}-3\hat{k} \\ \overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA}=\left(7\hat{i}+0\hat{j}+6\hat{k}\right)-\left(3\hat{i}\text{+4}\hat{j}\text{+2}\hat{k}\right)=4\hat{i}-4\hat{j}+4\hat{k} \\ \overrightarrow{n}=\overrightarrow{AB}\times \overrightarrow{AC}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -1& -6& -3\\ 4& -4& 4\end{array}\right|=-36\hat{i}-8\hat{j}+28\hat{k} \\ \text{The vector equation of the required plane is} \\ \overrightarrow{r}.\overrightarrow{n}=\overrightarrow{a}.\overrightarrow{n} \\ \Rightarrow \overrightarrow{r}.\left(-36\hat{i}-8\hat{j}+28\hat{k}\right)=\left(3\hat{i}\text{+4}\hat{j}\text{+2}\hat{k}\right).\left(-36\hat{i}-8\hat{j}+28\hat{k}\right) \\ \Rightarrow \overrightarrow{r}.\left[-4\left(9\hat{i}+2\hat{j}+7\hat{k}\right)\right]=-108-32+56 \\ \Rightarrow \overrightarrow{r}.\left[-4\left(9\hat{i}+2\hat{j}+7\hat{k}\right)\right]=-84 \\ \Rightarrow \overrightarrow{r}.\left(9\hat{i}+2\hat{j}+7\hat{k}\right)=21 \end{gathered}$$