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 is an IIT Kanpur graduate and has been teaching for 16+ years. At Teachoo, he breaks down Maths, Science and Computer Science into simple steps so students understand concepts deeply and score with confidence.
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