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

Transcript

Example 29 Show that the lines (๐‘ฅ โˆ’ ๐‘Ž + ๐‘‘)/(๐›ผ โˆ’ ๐›ฟ) = (๐‘ฆ โˆ’ ๐‘Ž)/๐›ผ = (๐‘ง โˆ’ ๐‘Ž โˆ’ ๐‘‘)/(๐›ผ + ๐›ฟ) nd (๐‘ฅ โˆ’ ๐‘ + ๐‘)/(๐›ฝ โˆ’ ๐›พ) = (๐‘ฆ โˆ’ ๐‘)/๐›ฝ = (๐‘ง โˆ’ ๐‘ โˆ’ ๐‘)/(๐›ฝ + ๐›พ) are coplanar.Two lines (๐‘ฅ โˆ’ ๐‘ฅ_1)/๐‘Ž_1 = (๐‘ฆ โˆ’ ๐‘ฆ_1)/๐‘_1 = (๐‘ง โˆ’ ๐‘ง_1)/๐‘_1 and (๐‘ฅ โˆ’ ๐‘ฅ_2)/๐‘Ž_2 = (๐‘ฆ โˆ’ ๐‘ฆ_2)/๐‘_2 = (๐‘ง โˆ’ ๐‘ง_2)/๐‘_2 are coplanar if |โ– 8(๐’™_๐Ÿโˆ’ ๐’™_๐Ÿ&๐’š_๐Ÿโˆ’๐’š_๐Ÿ&๐’›_๐Ÿโˆ’๐’›_๐Ÿ@๐’‚_๐Ÿ&๐’ƒ_๐Ÿ&๐’„_๐Ÿ@๐’‚_๐Ÿ&๐’ƒ_๐Ÿ&๐’„_๐Ÿ )| = 0 (๐’™ โˆ’ ๐’‚ + ๐’…)/(๐œถ โˆ’ ๐œน) = (๐’š โˆ’ ๐’‚)/๐œถ = (๐’› โˆ’ ๐’‚ โˆ’ ๐’…)/(๐œถ + ๐œน) (๐‘ฅ โˆ’ (๐‘Ž โˆ’ ๐‘‘))/(๐›ผ โˆ’ ๐›ฟ) = (๐‘ฆ โˆ’ ๐‘Ž)/๐›ผ = (๐‘ง โˆ’ (๐‘Ž + ๐‘‘))/(๐›ผ + ๐›ฟ) Comparing (๐‘ฅ โˆ’ ๐‘ฅ_1)/๐‘Ž_1 = (๐‘ฆ โˆ’ ๐‘ฆ_1)/๐‘_1 = (๐‘ง โˆ’ ๐‘ง_1)/๐‘_1 ๐‘ฅ_1 = ๐‘Ž โˆ’ d , ๐‘ฆ_1= ๐‘Ž , ๐‘ง_1= ๐‘Ž + d & ๐‘Ž_1=๐›ผโˆ’๐›ฟ, ๐‘_1= ๐›ผ, ๐‘_1= ๐›ผ+๐›ฟ (๐’™ โˆ’ ๐’ƒ + ๐’„)/(๐œท โˆ’ ๐œธ) = (๐’š โˆ’ ๐’ƒ)/๐œท = (๐’› โˆ’ ๐’ƒ โˆ’ ๐’„)/(๐œท + ๐œธ) (๐‘ฅ โˆ’ (๐‘ โˆ’ ๐‘))/(๐›ฝ โˆ’ ๐›พ) = (๐‘ฆ โˆ’ ๐‘)/๐›ฝ = (๐‘ง โˆ’ (๐‘ + ๐‘))/(๐›ฝ + ๐›พ) Comparing (๐‘ฅ โˆ’ ๐‘ฅ_2)/๐‘Ž_2 = (๐‘ฆ โˆ’ ๐‘ฆ_2)/๐‘_2 = (๐‘ง โˆ’ ๐‘ง_1)/๐‘_2 ๐‘ฅ_2 = ๐‘ โˆ’ c , ๐‘ฆ_2= ๐‘ , ๐‘ง_2= ๐‘ + c & ๐‘Ž_2 = ๐›ฝโˆ’๐›พ, ๐‘_2 = ๐›ฝ, ๐‘_2 = ๐›ฝ + ๐›พ Now, |โ– 8(๐‘ฅ_2โˆ’๐‘ฅ_1&๐‘ฆ_2โˆ’๐‘ฆ_1&๐‘ง_2โˆ’๐‘ง_1@๐‘Ž_1&๐‘_1&๐‘_1@๐‘Ž_2&๐‘_2&๐‘_2 )| = |โ– 8(๐‘โˆ’๐‘โˆ’๐‘Ž + ๐‘‘&๐‘โˆ’๐‘Ž&๐‘+๐‘โˆ’๐‘Žโˆ’๐‘‘@๐›ผโˆ’๐›ฟ&๐›ผ&๐›ผ+๐›ฟ@๐›ฝโˆ’๐›พ&๐›ฝ&๐›ฝ+๐›พ)| Adding Column 3 to Column 1, = |โ– 8(๐‘โˆ’๐‘โˆ’๐‘Ž + ๐‘‘+(๐‘+๐‘โˆ’๐‘Žโˆ’๐‘‘)&๐‘โˆ’๐‘Ž&๐‘+๐‘โˆ’๐‘Žโˆ’๐‘‘@๐›ผโˆ’๐›ฟ+(๐›ผ+๐›ฟ)&๐›ผ&๐›ผ+๐›ฟ@๐›ฝโˆ’๐›พ+(๐›ฝ+๐›พ)&๐›ฝ&๐›ฝ+๐›พ)| = |โ– 8(2(๐‘โˆ’๐‘Ž)&๐‘โˆ’๐‘Ž&๐‘+๐‘โˆ’๐‘Žโˆ’๐‘‘@2๐›ผ&๐›ผ&๐›ผ+๐›ฟ@2๐›ฝ&๐›ฝ&๐›ฝ+๐›พ)| Taking 2 common from Column 1 = 2 |โ– 8(๐‘ โˆ’ ๐‘Ž&๐‘ โˆ’ ๐‘Ž&๐‘ + ๐‘ โˆ’ ๐‘Ž โˆ’ ๐‘‘@๐›ผ&๐›ผ&๐›ผ + ๐›ฟ@๐›ฝ&๐›ฝ&๐›ฝ +๐›พ)| = 2 ร— 0 = 0 Therefore, the given two lines are coplanar. Since Columns 1 and 2 are same, The value of determinant is zero.

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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.