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

Transcript

Ex 11.3, 12 Find the angle between the planes whose vector equations are ๐‘Ÿ โƒ— . (2๐‘– ฬ‚ + 2๐‘— ฬ‚ โ€“ 3๐‘˜ ฬ‚) = 5 and ๐‘Ÿ โƒ— . (3๐‘– ฬ‚ โ€“ 3๐‘— ฬ‚ + 5๐‘˜ ฬ‚) = 3 .Angle between two planes ๐‘Ÿ โƒ— . (๐‘›_1 ) โƒ— = d1 and ๐‘Ÿ โƒ—.(๐‘›2) โƒ— = d2 is given by cos ๐œƒ = |((๐’๐Ÿ) โƒ—. (๐’๐Ÿ) โƒ—)/|(๐’๐Ÿ) โƒ— ||(๐’๐Ÿ) โƒ— | | ๐’“ โƒ—.(2๐’Š ฬ‚ + 2๐’‹ ฬ‚ โˆ’ 3๐’Œ ฬ‚) = 5 Comparing with ๐‘Ÿ โƒ—.(๐‘›1) โƒ— = (๐‘‘1) โƒ—, (๐‘›1) โƒ— = 2๐‘– ฬ‚ + 2๐‘— ฬ‚โˆ’3๐‘˜ ฬ‚ Magnitude of (๐‘›1) โƒ— = โˆš(2^2+2^2+ใ€–(โˆ’3)ใ€—^2 ) |(๐‘›1) โƒ— | = โˆš(4+4+9) = โˆš17 So, cos ฮธ = |((2๐‘– ฬ‚ + 2๐‘— ฬ‚ โˆ’ 3๐‘˜ ฬ‚ ) . (3๐‘– ฬ‚ โˆ’ 3๐‘— ฬ‚ + 5๐‘˜ ฬ‚ ))/(โˆš17 ร— โˆš43)| = |((2 ร— 3) + (2 ร— โˆ’3) + (โˆ’3 ร— 5))/โˆš(17 ร— 43)| = |(6 โˆ’ 6 โˆ’ 15)/โˆš731| = |(โˆ’15)/โˆš731| = 15/โˆš731 So, cos ฮธ = 15/โˆš731 โˆด ฮธ = cosโˆ’1(๐Ÿ๐Ÿ“/โˆš๐Ÿ•๐Ÿ‘๐Ÿ) Therefore, the angle between the planes is cosโˆ’1(15/โˆš731).

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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.