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/โ83 (d) 6
Here, the two planes are parallel
We know that
Distance between two parallel planes Ax + By + Cz = ๐_1 and Ax + By + Cz = ๐_2 is
|(๐ _๐ โ ๐ _๐)/(โ(๐จ^๐ + ๐ฉ^๐ + ๐ช^๐ ) )|
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 โ 3)/โ(3^2 + 5^2 + 7^2 )|
= ๐ }1 mark
Hence, (A) is the correct option

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.