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

The given planes are
$\overrightarrow{r}.(\hat{i}-2\hat{j}+3\hat{k})-4=0$ and
$\overrightarrow{r}(-2\hat{i}+\hat{j}+\hat{k})+5=0$

Changing to cartesian form, the equations of these planes are:
$x-2y+3z-4=0$ and
$-2x+y+z+5=0$

Equation of any plane through the intersection of these plnaes is:
$x-2y+3z-4+\lambda (-2x+y+z+5)=0....\left(1\right)$
For the intercept on $x$-axis, putting $y=0,z=0$, we have
$x-4+\lambda (-2x+5)=0$
$\Rightarrow x(1-2\lambda )=4-5\lambda$
$\Rightarrow x=\frac{4-5\lambda }{1-2\lambda }$

For the intercept on $y$-axis, putting $x=0,z=0$, we have
$-2y-4+\lambda y+5\lambda =0$
$(-2+\lambda )y=4-5\lambda$
$\Rightarrow y=\frac{4-5\lambda }{-2+\lambda }$

$\therefore$ Intercepts on $x$-axis and $y$ -axis made by plane
(1) are $\frac{4-5\lambda }{1-2\lambda }$ and $\frac{4-5\lambda }{-2+\lambda }$ respectively.

Also, it is given that intercepts on $x$-axis and $y$ -axis are equal
$\Rightarrow \frac{(4-5\lambda )}{1-2\lambda }=\frac{4-5\lambda }{-2+\lambda }$
$\Rightarrow (4-5\lambda )(-2+\lambda )=(4-5\lambda )(1-2\lambda )$
$\Rightarrow -2+\lambda =1-2\lambda$
$\Rightarrow \lambda =1$

Substituting $\lambda =1$, we have plane (1)
$x-2y+3z-4+1(-2x+y+z+5)=0$

$\Rightarrow -x-y+4z+1=0$

hence the required plane is
$\Rightarrow x+y-4z-1=0.$