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  1. Chapter 11 Class 12 Three Dimensional Geometry
  2. Serial order wise

Transcript

Misc 22 (Method 1) Distance between two planes: 2x + 3y + 4z = 4 and 4x + 6y + 8z = 12 is (A) 2 units (B) 4 units (C) 8 units (D) 2/โˆš29 units 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 ) )| 2x + 3y + 4z = 4 Comparing with Ax + By + Cz = d1 A = 2, B = 3, C = 4, d1 = 4 4x + 6y + 8z = 12 2 (2x + 3y + 4z) = 12 Dividing by 2 2x + 3y + 4z = 6 Comparing with Ax + By + Cz = d2 A = 2, B = 3, C = 4 , d2 = 6 So, Distance between the two planes = |(4 โˆ’ 6)/โˆš(2^2 + 3^2 + 4^2 )| = |(โˆ’2)/โˆš(4 + 9 + 16)| = ๐Ÿ/โˆš๐Ÿ๐Ÿ— Hence, (D) is the correct option Misc 22 (Method 2) Distance between the two planes : 2x + 3y + 4z = 4 and 4x + 6y + 8z = 12 is (A) 2 units (B) 4 units (C) 8 units (D)2/โˆš29 units Distance of a point (๐‘ฅ_1, ๐‘ฆ_1, ๐‘ง_1) from the plane Ax + By + Cz = D is |(๐‘จ๐’™_๐Ÿ + ๐‘ฉ๐’š_๐Ÿ + ๐‘ช๐’›_๐Ÿโˆ’ ๐‘ซ)/โˆš(๐‘จ^๐Ÿ + ๐‘ฉ^๐Ÿ + ๐‘ช^๐Ÿ )| Let us take a point P (๐‘ฅ_1, ๐‘ฆ_1, ๐‘ง_1) on the plane 2x + 3y + 4z = 4 2๐‘ฅ_1 + 3๐‘ฆ_1 + 4๐‘ง_1 = 4 Now, to find the distance of point P form plane 4x + 6y + 8z = 12, Comparing with Ax + By + Cz = D, A = 4, B = 6, C = 8, D = 12 Distance of P (๐‘ฅ_1, ๐‘ฆ_1, ๐‘ง_1) from the plane 4x + 6y + 8z = 12 = |(4๐‘ฅ_1+ ใ€–6๐‘ฆใ€—_1 + 8๐‘ง_1โˆ’ 12)/โˆš(4^2 + 6^2 + 8^2 )| = |(2(๐Ÿ๐’™_๐Ÿ + ใ€–๐Ÿ‘๐’šใ€—_๐Ÿ + ๐Ÿ’๐’›_๐Ÿ )โˆ’ 12)/โˆš(16 + 36 + 64)| = |(2 ร— ๐Ÿ’ โˆ’ 12)/โˆš116| = |(8 โˆ’ 12)/โˆš(4 ร— 29)| = |(โˆ’4)/(2โˆš29)| = ๐Ÿ/โˆš๐Ÿ๐Ÿ— Hence, (D) 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.