Q113 / 180 · KCET 2024 · Mathematics
The distance between the two planes $2 x+3 y+4 z=4$ and $4 x+6 y+8 z=12$ is
- A2 units
- B8 units
- Correct answerC$\frac{2}{\sqrt{29}}$ unit
- D4 units
Solution
The equations of planes are
$2 x+3 y+4 z=4$ ....(i)
$4 x+6 y+8 z=12$
$2 x+3 y+4 z=6$ ....(ii)
It can be seen that the given planes are parallel. It is known that the distance between two parallel
planes is $\left|\frac{d_2-d_1}{\sqrt{a^2+b^2+c^2}}\right|$.
So, $\quad D=\left|\frac{6-4}{\sqrt{(2)^2+(3)^2+(4)^2}}\right|=\frac{2}{\sqrt{29}}$ units
Hence, the correct answer is option: 'C'.