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/√83 (d) 6
Here, the two planes are parallel
We know that
Distance between two parallel planes Ax + By + Cz = 𝑑_1 and Ax + By + Cz = 𝑑_2 is
|(𝒅_𝟏 − 𝒅_𝟐)/(√(𝑨^𝟐 + 𝑩^𝟐 + 𝑪^𝟐 ) )|
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 − 3)/√(3^2 + 5^2 + 7^2 )|
= 𝟎 }1 mark
Hence, (A) is the correct option
Made by
Davneet Singh
Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo
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