Question 9 - CBSE Class 12 Sample Paper for 2020 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards
Last updated at Oct. 22, 2019 by Teachoo

What is the distance(in units) between the two planes
3x + 5y + 7z = 3 and 9x + 15y + 21z = 9 ?
(a) 0Β Β Β Β Β Β Β (b) 3Β Β Β Β Β Β Β (c) 6/β83Β Β Β Β Β Β Β Β (d) 6

Note
: - This
is similar to
Misc
22 of NCERT β Chapter 11 Class 12

Check the answer here
https://www.teachoo.com/3555/755/Misc-22---Distance-between-two-planes---2x---3y---4z--4/category/Miscellaneous/

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Transcript
Question 9 What is the distance(in units) between the two planes 3x + 5y + 7z = 3 and 9x + 15y + 21z = 9 ? (a) 0 (b) 3 (c) 6/β83 (d) 6
Here, the two planes are parallel
We know that
Distance between two parallel planes Ax + By + Cz = π_1 and Ax + By + Cz = π_2 is
|(π
_π β π
_π)/(β(π¨^π + π©^π + πͺ^π ) )|
3x + 5y + 7z = 3
Comparing with Ax + By + Cz = d1
A = 3, B = 5, C = 7, d1 = 3
9x + 15y + 21z = 9
3(3x + 5y + 7z) = 9
Dividing by 3
3x + 5y + 7z = 3
Comparing with Ax + By + Cz = d2 A = 3, B = 5, C = 7 , d2 = 3
So,
Distance between the two planes
= |(3 β 3)/β(3^2 + 5^2 + 7^2 )|
= π }1 mark
Hence, (A) is the correct option

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