Question

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

Answer 1

$\overrightarrow{r}.(\hat{i}+2\hat{j}+3\hat{k})-4=\mathrm{0.....}\left(1\right)$

$\overrightarrow{r}.(2\hat{i}+\hat{j}-\hat{k})+5=\mathrm{0......}\left(2\right)$

The equations of the plane passing through the line intersection of the plane given in equation (1) and equation (2) is
$[\overrightarrow{r}.(\hat{i}+2\hat{j}+3\hat{k})-4]+\lambda [\overrightarrow{r}.(2\hat{i}+\hat{j}-\hat{k})+5]=0$

$\overrightarrow{r}.\left[\right(2\lambda +1)\hat{i}+(\lambda +2)\hat{j}+(3-\lambda \left)\hat{k}\right]+(5\lambda -4)=\mathrm{0.....}\left(3\right)$

The plane in equation (3) is perpendicular to the plane

$\therefore$ $5(2\lambda +1)+3(\lambda +2)-6(3-\lambda )=0$

$\Rightarrow$ $19\lambda -7=0$

$\Rightarrow$ $\lambda =\frac{7}{19}$

Substituting $\lambda =\frac{7}{19}$ in equation (3) we obtain

$\Rightarrow$ $\overrightarrow{r}.[\frac{33}{19}\hat{i}+\frac{45}{19}\hat{j}+\frac{50}{19}\hat{k}]-\frac{41}{19}=0$

$\Rightarrow$ $\overrightarrow{r}.[(33\hat{i}+45\hat{j}+50\hat{k})-41]=\mathrm{0.......}\left(4\right)$

This is the vector equation of the required plane.

The cartesian equation of this plane can be obtained by substituting $\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}$ in equation (3).

$(x\hat{i}+y\hat{j}+z\hat{k}).(33\hat{i}+45\hat{j}+50\hat{k})-41=0$

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