Question:

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

Solution:

The equations of the given planes are

$\vec{r} \cdot(\hat{i}+2 \hat{j}+3 \hat{k})-4=0$

$\vec{r} \cdot(2 \hat{i}+\hat{j}-\hat{k})+5=0$

The equation of the plane passing through the line intersection of the plane given in equation (1) and equation (2) is

$[\vec{r} \cdot(\hat{i}+2 \hat{j}+3 \hat{k})-4]+\lambda[\vec{r} \cdot(2 \hat{i}+\hat{j}-\hat{k})+5]=0$

$\vec{r} \cdot[(2 \lambda+1) \hat{i}+(\lambda+2) \hat{j}+(3-\lambda) \hat{k}]+(5 \lambda-4)=0$                         $\ldots(3)$

The plane in equation $(3)$ is perpendicular to the plane, $\vec{r} \cdot(5 \hat{i}+3 \hat{j}-6 \hat{k})+8=0$

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

$\Rightarrow 19 \lambda-7=0$

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

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

$\Rightarrow \vec{r} \cdot\left[\frac{33}{19} \hat{i}+\frac{45}{19} \hat{j}+\frac{50}{19} \hat{k}\right] \frac{-41}{19}=0$

$\Rightarrow \vec{r} \cdot(33 \hat{i}+45 \hat{j}+50 \hat{k})-41=0$                   $\ldots(4)$

This is the vector equation of the required plane.

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

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

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