Question

The equation of line of intersection of planes $4x+4y-5z=12,8x+12y-13z=32$ can be written as :

• A. $\frac{x-1}{2}=\frac{y+2}{-3}=\frac{z}{4}$
• B. $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z}{4}$
• C. $\frac{x}{2}=\frac{y+2}{3}=\frac{z-2}{4}$
• D. $\frac{x}{2}=\frac{y}{3}=\frac{z-2}{4}$

Answer 1

Answer: (B)

Given that, $4x+4y-5z=12...(i)$
and $8x+12y-13z=32...(ii)$
we know direction ratio's of the line are
$(x,y,z)=(l,m,n)$
Eq. $(i)$ and $(ii)$ becomes
$4l+4m-5n=0...(iii)$
and $8l+12m-13n=0...(iv)$
or $\frac{l}{8}=\frac{m}{12}=\frac{n}{16}$
$\Rightarrow \frac{l}{2}=\frac{m}{3}=\frac{n}{4}$
Intersection point with $z=0$ is given by
$4x+4y=12...(v)$
$\therefore 8x+12y=32...(vi)$
On solving Eqs. $(v)$ and $(vi)$, we get $(1,2,0)$.
$\therefore$ Required line is $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z}{4}$.