Question

Find the equation of the plane that contains the line of intersection of the planes $\overrightarrow{r}·\left(\hat{i}+2\hat{j}+3\hat{k}\right)-4=0\mathrm{and}\overrightarrow{r}·\left(2\hat{i}+\hat{j}-\hat{k}\right)+5=0$ and which is perpendicular to the plane $\overrightarrow{r}·\left(5\hat{i}+3\hat{j}-6\hat{k}\right)+8=0.$

Answer 1

$$\begin{gathered} \text{The equation of the plane passing through the line of intersection of the given planes is} \\ \overrightarrow{r}.\left(\hat{i}+2\hat{j}+3\hat{k}\right)-4+\lambda \left[\overrightarrow{r}.\left(2\hat{i}+\hat{j}-\hat{k}\right)+5\right]=0 \\ \overrightarrow{r}.\left[\left(1+2\lambda \right)\hat{i}+\left(2+\lambda \right)\hat{j}+\left(3-\lambda \right)\hat{k}\right]-4+5\lambda =0...\left(1\right) \\ \text{This plane is perpendicular to}\overrightarrow{r}.\left(5\hat{i}+3\hat{j}-6\hat{k}\right)+8=0.\text{So,} \\ 5\left(1+2\lambda \right)+3\left(2+\lambda \right)-6\left(3-\lambda \right)=0\text{(Because}{a}_1{a}_2+b_1{b}_2+c_1{c}_2=0\text{)} \\ \Rightarrow 5+10\lambda +6+3\lambda -18+6\lambda =0 \\ \Rightarrow 19\lambda -7=0 \\ \Rightarrow \lambda =\frac{7}{19} \\ \text{Substituting this in (1), we get} \\ \overrightarrow{r}.\left[\left(1+2\left(\frac{7}{19}\right)\right)\hat{i}+\left(2+\frac{7}{19}\right)\hat{j}+\left(3-\frac{7}{19}\right)\hat{k}\right]-4+5\left(\frac{7}{19}\right)=0 \\ \Rightarrow \overrightarrow{r}.\left(33\hat{i}+45\hat{j}+50\hat{k}\right)-41=0 \\ \Rightarrow \left(x\hat{i}+y\hat{j}+z\hat{k}\right).\left(33\hat{i}+45\hat{j}+50\hat{k}\right)-41=0 \\ \Rightarrow 33x+45y+50z-41=0 \end{gathered}$$