Question

Find the vector and Cartesian equations of the plane passing through the line of intersection of the planes $\overrightarrow{r}\cdot (2\hat{i}+2\hat{j}-3\hat{k})=7,\overrightarrow{r}\cdot (2\hat{i}+5\hat{j}+3\hat{k})=9$ such that the intercepts made by the plane on x-axis and z-axis are equal.

Answer 1

Given Planes (Cartesian form)

$2x+2y-3z-7=0$

and $2x+3y+3z-9=0$

Plane passing through their point of intersection

$\Rightarrow (2x+2y-3z-7)+\lambda (2x++5y+3z-9)=0$

$\Rightarrow x(2+2\lambda )+y(2+5\lambda )+z(-3+3\lambda )+(-9\lambda -7)=0$

$\Rightarrow x\left(\frac{2+2\lambda }{9\lambda +7}\right)+y\left(\frac{2+5\lambda }{9\lambda +7}\right)+z\left(\frac{3\lambda -3}{9\lambda +7}\right)=1$

$\because xandz$ intercepts are equal

$\Rightarrow \frac{2+2\lambda }{9\lambda +7}=\frac{3\lambda -3}{9\lambda +7}$

$\Rightarrow 2+2\lambda =3\lambda -3$

$\Rightarrow \lambda =5$

$\therefore$ Required equation of plane

$\Rightarrow x\left(\frac{12}{52}\right)+y\left(\frac{27}{52}\right)+z\left(\frac{12}{52}\right)=1$

$\Rightarrow 12x+27y+12z=52$

$\Rightarrow 4x+9y+4z=14$

Vector form

$\Rightarrow \overrightarrow{r}.(4\overrightarrow{i}+9\overrightarrow{j}+4\overrightarrow{k})-14=0$