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

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

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

$\Rightarrow x-2y+3z-4=0\&-2x+y+z+5=0$

Any plane passing through the line of intersect.

$x-2y+3z-4+\lambda (-2x+y+z+5)=0$

$x(1-2\lambda )+y(-2+\lambda )+z(3+\lambda )+(5\lambda -4)=0$

Intercepts are equal on axe

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

$\Rightarrow -2+\lambda =1-2\lambda$

$\Rightarrow 3\lambda =3\Rightarrow \lambda =1$

Therefore, the required plane is $-x-y+4z+1=0orx+y-4z-1=0.$