22.   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) $\dpi{80} \frac{2}{\sqrt{29}}unit$

Given equations are
$2x+3y+4z= 4 \ \ \ \ \ \ \ \ \ -(i)$
and
$4x+6y+8z= 12\\ 2(2x+3y+4z)= 12\\ 2x+3y+4z = 6 \ \ \ \ \ \ \ \ \ \ -(ii)$
Now, it is clear from  equation (i) and (ii) that given planes are parallel
We know that the distance between two parallel planes $ax+by +cz = d_1 \ and \ ax+by +cz = d_2$  is given by
$D= \left | \frac{d_2-d_1}{\sqrt{a^2+b^2+c^2}} \right |$
Put the values in this equation
we will get,
$D= \left | \frac{6-4}{\sqrt{2^2+3^2+4^2}} \right |$
$D= \left | \frac{2}{\sqrt{4+9+16}} \right |= \left | \frac{2}{\sqrt{29}} \right |$
Therefore, the correct answer is (D)

Related Chapters

Preparation Products

Knockout NEET May 2021

An exhaustive E-learning program for the complete preparation of NEET..

₹ 22999/- ₹ 14999/-
Knockout BITSAT 2020

It is an exhaustive preparation module made exclusively for cracking BITSAT..

₹ 4999/- ₹ 1999/-
Knockout NEET May 2022

An exhaustive E-learning program for the complete preparation of NEET..

₹ 34999/- ₹ 24999/-
Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 22999/- ₹ 14999/-