Question

Consider two planes $P_1:2x-2y-z=4$ and $P_2:x+3y+\frac{5}{2}z=6,$ then a plane passing through the intersection of planes $P_1$ and $P_2$ and is equally inclined to coordinate axes is ?

• A. $x-y-z=2$
• B. $x-y+z=8$
• C. $x+y+z=4$
• D. $x+y-z=6$

Answer 1

The correct option is C $x+y+z=4$

Given planes, $P_1:2x-2y-z=4$ and $P_2:x+3y+\frac{5}{2}z=6,$
Plane passing through intercetion of $P_1,P_2$ is:
$P_1+\lambda P_2=0\Rightarrow 2x-2y-z-4+\lambda (x+3y+\frac{5}{2}z-6)=0\Rightarrow (\lambda +2)x+(3\lambda -2)y+(-1+\frac{5}{2}\lambda )z-4-6\lambda =0$
So, intercepts are : $\frac{6\lambda +4}{\lambda +2},\frac{6\lambda +4}{3\lambda -2},\frac{6\lambda +4}{\frac{5}{2}\lambda -1}$
$\Rightarrow \frac{6\lambda +4}{\lambda +2}=\frac{6\lambda +4}{3\lambda -2}=\frac{6\lambda +4}{\frac{5}{2}\lambda -1}\Rightarrow \lambda +2=3\lambda -2=\frac{5}{2}\lambda -1\Rightarrow \lambda =2$
So, equation of plane is:
$2x-2y-z-4+2(x+3y+\frac{5}{2}z-6)=0\Rightarrow 4x+4y+4z-16=0\Rightarrow x+y+z=4$