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

What is the distance between planes  3x + 5y + 7z = 3 and 9x+15y+21z=9

Question 9 - CBSE Class 12 Sample Paper for 2020 Boards - Part 2

Note : - This is similar to Misc 22 of NCERT – 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/√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

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.