Question

Find the equation of the plane which contains the line of intersection of the planes

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

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

and whose intercept on X-axis is equal to that of on Y-axis.

Answer 1

Given equation of planes are

$\overrightarrow{r}.(\hat{i}-2\hat{j}+3\hat{k})=4$

and

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

On comparing these with $\overrightarrow{r}.\overrightarrow{n}=d$,

we get,

${\overrightarrow{n}}_1=\hat{i}-2\hat{j}+3\hat{k},d_1=4$,

${\overrightarrow{n}}_2=-2\hat{i}+\hat{j}+\hat{k},d_2=-5$

Now, the equation of the plane which contains the intersection of the given planes is

$\overrightarrow{r}.({\overrightarrow{n}}_1+\lambda {\overrightarrow{n}}_2)=d_1+\lambda d_2$

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

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

Also, given intercept on X and Y-axes are same. Therefore,

$\frac{4-5\lambda }{1-2\lambda }=\frac{4-5\lambda }{-2+\lambda }$

$\lambda =1$

Putting this value in equation 1, we get,

$\overrightarrow{r}.(-\hat{i}-\hat{j}+4\hat{k})=-1$,

which is the required equation of the plane.