Question

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

Answer 1

We have,

$\begin{array}{c}\overrightarrow{r}.\left(\hat{i}+2\hat{j}+3\hat{k}\right)=4\overrightarrow{r}.\left(2\hat{i}+\hat{j}-\hat{k}\right)+5=0\\ x+2y+3z=42x+y-z=-5\\ {\pi }_3=\left(x+2y+3z-4\right)+\lambda \left(2x+y-z+5\right)=0\\ {\pi }_3\perp {\pi }_4{\pi }_4=\overrightarrow{r}.\left(5\hat{i}+3\hat{j}-6\hat{k}\right)+8=0\\ {\pi }_4=5x+3y-6z+8=0\\ {\pi }_3=x\left(1+2\lambda \right)+y\left(2+\lambda \right)+z\left(3-\lambda \right)-4+5\lambda =0\\ 5\left(1+2\lambda \right)+3\left(\lambda +2\right)-6\left(3-\lambda \right)=0\\ 11-18+10\lambda +3\lambda +6\lambda =0\\ 19\lambda =7\\ \lambda =\frac{7}{19}\\ x\left(1+\frac{14}{19}\right)+y\left(2+\frac{7}{19}\right)+z\left(3-\frac{7}{19}\right)-4+\frac{35}{19}=0\end{array}$

Hence equation of plane;

$33x+45y+50z-41=0$