Download MARKS App - Trusted by 15,00,000+ IIT JEE & NEET aspirants! Download Now
Search any question & find its solution
Question: Answered & Verified by Expert
22. Distance between the two planes: $2 x+3 y+4 z=4$ and $4 x+6 y+8 z=12$ is
(a) 2 units
(b) 4 units
(c) 8 units
(d) $\frac{2}{\sqrt{29}}$ units
MathematicsThree Dimensional Geometry
Solution:
2476 Upvotes Verified Answer
$2 x+3 y+4 z=6$ the distance between parallel planes
$a x+b y+c z=d_1$ and $a x+b y+c z=d_2$
$\therefore\left|\frac{\mathrm{d}_1-\mathrm{d}_2}{\sqrt{\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2}}\right|$
The distance between the planes (i) and (iii) is
$$
\left|\frac{6-4}{\sqrt{2^2+3^2+4^2}}\right|=\left|\frac{2}{\sqrt{4+9+16}}\right|=\frac{2}{\sqrt{29}}
$$
Option (d) is correct.
$2 x+3 y+4 z=6$ the distance between parallel planes
$a x+b y+c z=d_1$ and $a x+b y+c z=d_2$
$\therefore\left|\frac{\mathrm{d}_1-\mathrm{d}_2}{\sqrt{\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2}}\right|$
The distance between the planes (i) and (iii) is
$$
\left|\frac{6-4}{\sqrt{2^2+3^2+4^2}}\right|=\left|\frac{2}{\sqrt{4+9+16}}\right|=\frac{2}{\sqrt{29}}
$$
Option (d) is correct.

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.