Question

Find the equation of the plane through the intersection of the planes $\overrightarrow{r}.(\hat{i}+3\hat{j})-6=0$ and $\overrightarrow{r}.(3\hat{i}-\hat{j})-4\hat{k}=0$, whose perpendicular distance from origin is unity.

Answer 1

We have, $\overrightarrow{n_1}=(\hat{i}+3\hat{j}),d_1=6and\overrightarrow{n_2}=(3\hat{i}-\hat{j}-4\hat{k}),d_2=0$

Using the relation, $\overrightarrow{r}.(\overrightarrow{n_1}+\lambda \overrightarrow{n_2})=d_1+d_2\lambda \Rightarrow \overrightarrow{r}.\left[\right(\hat{i}+3\hat{j}+\lambda (3\hat{i}-\hat{j}-4\hat{k})]=6+0.\lambda \overrightarrow{r}.[(1+3\lambda )\hat{i}+(3-\lambda )\hat{j}+k(-4\lambda )]=6...(i)Ondividingbothsidesby\sqrt{(1+3\lambda {)}^2+(3-\lambda {)}^2+(-4\lambda {)}^2},weget\frac{\overrightarrow{r}.\left[\right(1+3\lambda )\hat{i}+(3-\lambda )\hat{j}+k(-4\lambda \left)\right]}{\sqrt{(1+3\lambda {)}^2+(3-\lambda {)}^2+(-4\lambda {)}^2}}=\frac{6}{\sqrt{(1+3\lambda {)}^2+(3-\lambda {)}^2+(-4\lambda {)}^2}}$

$\therefore \frac{6}{\sqrt{(1+3\lambda {)}^2+(3-\lambda {)}^2+(-4\lambda {)}^2}}=1\Rightarrow (1+3\lambda {)}^2+(3-\lambda {)}^2+(-4\lambda {)}^2=36\Rightarrow 1+9{\lambda }^2+6\lambda +9+{\lambda }^2-6\lambda +16{\lambda }^2=36\Rightarrow 26{\lambda }^2+10=36\Rightarrow {\lambda }^2=1\therefore \lambda =\pm 1$

Using Eq.(i),the required equation of plane is

$\overrightarrow{r}.\left[\right(1\pm 3)\hat{i}+(3\pm 1)\hat{j}+(\pm 4\left)\hat{j}\right]=6\Rightarrow \overrightarrow{r}.\left[\right(1+3)\hat{i}+(3-1)\hat{j}+(-4\left)\hat{j}\right]=6and\overrightarrow{r}.\left[\right(1-3)\hat{i}+(3+1)\hat{j}+4\hat{j}]=6\Rightarrow \overrightarrow{r}.(4\hat{i}+2\hat{j}-4\hat{k})=6and\overrightarrow{r}.(-2\hat{i}+4\hat{j}+4\hat{k})=6\Rightarrow 4x+2y-4z-6=0and-2x+4y+4z-6=0$