Question

Choose the correct answer in Question
Distance between the two planes 2x + 3y + 4z = 4 and 4x + 6y+ 8z =12 is
(a) 2 units
(b) 4 units
(c) 8 units
(d) $\frac{2}{\sqrt{29}}units$

Answer 1

Given planes are 2x + 3y + 4z =4 . . . (i)
and 4x + 6y + 8z =12 $\Rightarrow$ 2x + 3y + 4z =6 . . . (ii)
From the above equations, it can be seen that given planes are parallel. It is known that the distance between two planes, ax + by + cz = $d_1$ and ax + by + cz = $d_2$ is given by
$\therefore Distanced=\left|\frac{d_2-d_1}{\sqrt{a^2+b^2+c^2}}\right|\Rightarrow d=\left|\frac{6-4}{\sqrt{2^2+3^2+4^2}}\right|$
$\Rightarrow d=\frac{2}{\sqrt{4+9+16}}=\frac{2}{\sqrt{29}}$ units. So, the correct option is (d).