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

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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/

 

  1. Class 12
  2. Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

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/โˆš83 (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Ÿ โˆ’ ๐’…_๐Ÿ)/(โˆš(๐‘จ^๐Ÿ + ๐‘ฉ^๐Ÿ + ๐‘ช^๐Ÿ ) )| 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 โˆ’ 3)/โˆš(3^2 + 5^2 + 7^2 )| = ๐ŸŽ }1 mark Hence, (A) is the correct option

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Davneet Singh
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.