Question

If the equation of the line passing through $M(1,1,1)$ and intersecting at right angle to the line of intersection of the planes $x+2y-4z=0$ and $2x-y+2z=0$ is $\frac{x-1}{a}=\frac{y-1}{b}=\frac{z-1}{c},$ then $a:b:c$ equals

• A. $5:-1:2$
• B. $-5:1:2$
• C. $5:1:-2$
• D. $5:1:2$

Answer 1

The correct option is A $5:-1:2$

Given: $x+2y-4z=0$ and $2x-y+2z=0$
Since $(0,0,0)$ lies on both planes,
$\therefore$ Equation of the line of intersection of the these two planes is $\frac{x-0}{x_1}=\frac{y-0}{y_1}=\frac{z-0}{z_1}$
and $x_1+2y_1-4z_1=0\cdots \left(1\right)$
$2x_1-y_1+2z_1=0\cdots \left(2\right)$
From $\left(1\right)$ and $\left(2\right),$ we have
$\frac{x_1}{0}=\frac{y_1}{-10}=\frac{z_1}{-5}$
$\therefore \frac{x}{0}=\frac{y}{-10}=\frac{z}{-5}\cdots \left(3\right)$
Any general point on line $\left(3\right)$ is $P(0,-10\lambda ,-5\lambda )$



Now, direction ratios of the line joining $P$ and $M$ is $⟨1,1+10\lambda ,1+5\lambda ⟩$
As line $MP$ is perpendicular to line $\left(3\right),$
$\therefore 0\left(1\right)-10(1+10\lambda )-5(1+5\lambda )=0$
$\Rightarrow \lambda =-\frac{3}{25}$
So, direction ratios of line are $⟨1,-\frac{1}{5},\frac{2}{5}⟩$ or $<5,-1,2>$
Hence, equation of required line is $\frac{x-1}{5}=\frac{y-1}{-1}=\frac{y-1}{2}$