Question

If the distance between the plane $ax-4y+2z=k$ and the plane containing the lines $\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}\text{and}\frac{x-3}{4}=\frac{y-4}{5}=\frac{z-5}{6}\text{is,}2\sqrt{6}\text{then}\frac{|k|}{3}$ is equal to

Answer 1

Distance between two planes = distance of the point (2, 3, 4) from the plane $2x-4y+2z=k$.

Let l, m, n be d.c.'s of normal to the plane of given lines.

$\therefore l.3+k.4+n.5=0$

$l.4+m.5+n.6=0$

$\Rightarrow \frac{l}{24-25}=\frac{-m}{18-20}=\frac{n}{15-16}$

$\Rightarrow \frac{l}{-1}=\frac{m}{2}=\frac{n}{-1}$

But the plane $ax-4y+z=k$ must be parallel to this plane.

$\therefore \frac{a}{-1}=\frac{-4}{2}=\frac{1}{-1}\Rightarrow a=2$

$\therefore$ distance between the planes $=2\sqrt{6}$

= distance of the planes $2x-4y+2z-k=0$ from the point (2, 3, 4)

$=\left|\frac{4-12+8-k}{\sqrt{4+16+4}}\right|=\frac{|k|}{2\sqrt{6}}$

$\therefore 4\times 6=|k|$

$\Rightarrow \frac{|k|}{3}=\frac{24}{3}=8$