Question

Find the vector equation of the plane passing through three points with position vectors $\hat{i}+\hat{j}-2\hat{k},2\hat{i}-\hat{j}+\hat{k}\mathrm{and}\hat{i}+2\hat{j}+\hat{k}.$ Also, find the coordinates of the point of intersection of this plane and the line $\overrightarrow{r}=3\hat{i}-\hat{j}-\hat{k}+\lambda \left(2\hat{i}-2\hat{j}+\hat{k}\right).$

Answer 1

$$\begin{gathered} \text{Let}\mathit{\text{A}}\text{(1, 1 , -2),}\mathit{\text{B}}\text{(2, -1, 1) and}\mathit{\text{C}}\text{(1, 2, 1) be the points represented by the given position vectors.} \\ \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{- 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}-\hat{j}+\hat{k}\right)-\left(\hat{i}\text{+}\hat{j}\text{- 2}\hat{k}\right)=\hat{i}-2\hat{j}+3\hat{k} \\ \overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA}=\left(\hat{i}+2\hat{j}+\hat{k}\right)-\left(\hat{i}\text{+}\hat{j}\text{-2}\hat{k}\right)=0\hat{i}+\hat{j}+3\hat{k} \\ \overrightarrow{n}=\overrightarrow{AB}\times \overrightarrow{AC}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -2& 3\\ 0& 1& 3\end{array}\right|=-9\hat{i}-3\hat{j}+\hat{k} \\ \text{The vector equation of the required plane is} \\ \overrightarrow{r}.\overrightarrow{n}=\overrightarrow{a}.\overrightarrow{n} \\ \Rightarrow \overrightarrow{r}.\left(-9\hat{i}-3\hat{j}+\hat{k}\right)=\left(\hat{i}\text{+}\hat{j}\text{-2}\hat{k}\right).\left(-9\hat{i}-3\hat{j}+\hat{k}\right) \\ \Rightarrow \overrightarrow{r}.\left(-9\hat{i}-3\hat{j}+\hat{k}\right)=-9-3-2 \\ \Rightarrow \overrightarrow{r}.\left[-\left(9\hat{i}+3\hat{j}-\hat{k}\right)\right]=-14 \\ \Rightarrow \overrightarrow{r}.\left(9\hat{i}+3\hat{j}-\hat{k}\right)=14 \\ \mathbf{To}\mathbf{f}\mathbf{\text{ind the point of intersection of this plane}} \\ \text{The given equation of the line is} \\ \overrightarrow{r}=\left(3\hat{i}-\hat{j}-\hat{k}\right)+\lambda \left(2\hat{i}-2\hat{j}+\hat{k}\right) \\ \Rightarrow \overrightarrow{r}=\left(3+2\lambda \right)\hat{i}+\left(-1-2\lambda \right)\hat{j}+\left(-1+\lambda \right)\hat{k} \\ \text{T}\mathrm{he\; coordinates\; of\; any\; point\; on\; this\; line\; are}\mathrm{in}the\; for\mathrm{m}\mathrm{of}\left(3+2\lambda \right)\hat{i}+\left(-1-2\lambda \right)\hat{j}+\left(-1+\lambda \right)\hat{k}\text{or}\left(3+2\lambda ,-1-2\lambda ,-1+\lambda \right) \\ \text{Since this point lies on the plane}\overrightarrow{r}\text{.}\left(9\hat{i}+3\hat{j}-\hat{k}\right)=14, \\ \left[\left(3+2\lambda \right)\hat{i}+\left(-1-2\lambda \right)\hat{j}+\left(-1+\lambda \right)\hat{k}\right].\left(9\hat{i}+3\hat{j}-\hat{k}\right)=14 \\ \Rightarrow 27+18\lambda -3-6\lambda +1-\lambda =14 \\ \Rightarrow 11\lambda =-11 \\ \Rightarrow \lambda =-1 \\ \mathrm{So,\; the\; coordinates\; of\; the\; point}\text{are} \\ \left(3+2\lambda ,-1-2\lambda ,-1+\lambda \right) \\ =\left(3-2,-1+2,-1-1\right) \\ =\left(1,1,-2\right) \end{gathered}$$