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

Transcript

Misc 6 If the lines (๐‘ฅ โˆ’ 1)/( โˆ’ 3) = (๐‘ฆ โˆ’ 2)/2๐‘˜ = (๐‘ง โˆ’ 3)/2 and (๐‘ฅ โˆ’ 1)/3๐‘˜ = (๐‘ฆ โˆ’ 1)/1 = (๐‘ง โˆ’ 6)/( โˆ’ 5) are perpendicular, find the value of k. Two lines (๐‘ฅ โˆ’ ๐‘ฅ1)/( ๐‘Ž1) = (๐‘ฆ โˆ’๐‘ฆ1)/๐‘1 = (๐‘ง โˆ’ ๐‘ง1)/๐‘1 and (๐‘ฅ โˆ’ ๐‘ฅ2)/( ๐‘Ž2) = (๐‘ฆ โˆ’ ๐‘ฆ2)/๐‘2 = (๐‘ง โˆ’ ๐‘ง2)/๐‘2 are perpendicular to each other if ๐’‚๐Ÿ ๐’‚๐Ÿ + ๐’ƒ๐Ÿ ๐’ƒ๐Ÿ + ๐’„๐Ÿ ๐’„๐Ÿ = 0 (๐’™ โˆ’ ๐Ÿ)/( โˆ’ ๐Ÿ‘) = (๐’š โˆ’ ๐Ÿ)/( ๐Ÿ๐’Œ) = (๐’› โˆ’ ๐Ÿ‘)/( ๐Ÿ) Comparing with (๐‘ฅ โˆ’ ๐‘ฅ1)/( ๐‘Ž1) = (๐‘ฆโˆ’ ๐‘ฆ1)/( ๐‘1) = (๐‘งโˆ’ ๐‘ง1)/( ๐‘1) x1 = 1, y1 = 2, z1 = 3 & ๐’‚๐Ÿ = โˆ’3, b1 = 2k c1 = 2 (๐’™ โˆ’ ๐Ÿ)/( ๐Ÿ‘๐’Œ) = (๐’š โˆ’ ๐Ÿ)/( ๐Ÿ) = (๐’› โˆ’ ๐Ÿ”)/( โˆ’ ๐Ÿ“) Comparing with (๐‘ฅ โˆ’ ๐‘ฅ2)/( ๐‘Ž2) = (๐‘ฆ โˆ’ ๐‘ฆ2)/( ๐‘2) = (๐‘ง โˆ’ ๐‘ง2)/( ๐‘2), ๐‘ฅ2 = 1, y2 = 1, z2 = 6 & ๐’‚๐Ÿ = 3k, b2 = 1, c2 = โˆ’5, Since the two lines are perpendicular, ๐‘Ž1 ๐‘Ž2 + ๐‘1 ๐‘2 + c1๐‘2 = 0 (โˆ’3 ร— 3k) + (2k ร— 1) + (2 ร— โˆ’ 5) = 0 โˆ’ 9k + 2k โˆ’ 10 = 0 โˆ’ 7k = 10 k = (โˆ’๐Ÿ๐ŸŽ)/๐Ÿ• Therefore, k = (โˆ’10)/7 Misc 7 Find the vector equation of the line passing through (1, 2, 3) and perpendicular to the plane ๐‘Ÿ โƒ—.(๐‘– ฬ‚ + 2๐‘— ฬ‚ โˆ’ 5๐‘˜ ฬ‚) + 9 = 0 . The vector equation of a line passing through a point with position vector ๐‘Ž โƒ— and parallel to vector ๐‘ โƒ— is ๐’“ โƒ— = ๐’‚ โƒ— + ๐œ†๐’ƒ โƒ— Given, the line passes (1, 2, 3) So, ๐’‚ โƒ— = 1๐’Š ฬ‚ + 2๐’‹ ฬ‚ + 3๐’Œ ฬ‚ Finding normal of plane ๐‘Ÿ โƒ—.(๐‘– ฬ‚ + 2๐‘— ฬ‚ โˆ’ 5๐‘˜ ฬ‚) + 9 = 0 ๐‘Ÿ โƒ—. (๐‘– ฬ‚ + 2๐‘— ฬ‚ โˆ’ 5๐‘˜ ฬ‚) = โˆ’ 9 โˆ’๐‘Ÿ โƒ—. (๐‘– ฬ‚ + 2๐‘— ฬ‚ โˆ’ 5๐‘˜ ฬ‚) = 9 ๐‘Ÿ โƒ—. (โˆ’1๐‘– ฬ‚ โˆ’ 2๐‘— ฬ‚ + 5๐‘˜ ฬ‚) = 9 Comparing with ๐‘Ÿ โƒ—. ๐‘› โƒ— = d, ๐’ โƒ— = โˆ’๐‘– ฬ‚ โˆ’ 2๐‘— ฬ‚ + 5๐‘˜ ฬ‚ Since line is perpendicular to plane, the line will be parallel to the normal of the plane โˆด ๐’ƒ โƒ— = ๐’ โƒ— = โˆ’1๐‘– ฬ‚ โˆ’ 2๐‘– ฬ‚ + 5๐‘˜ ฬ‚ Hence, ๐‘Ÿ โƒ— = (1๐‘– ฬ‚ + 2๐‘– ฬ‚ + 3๐‘˜ ฬ‚) + ๐œ†(โˆ’1๐‘– ฬ‚ โˆ’ 2๐‘– ฬ‚ + 5๐‘˜ ฬ‚) ๐‘Ÿ โƒ— = (1๐‘– ฬ‚ + 2๐‘– ฬ‚ + 3๐‘˜ ฬ‚) โˆ’ ๐œ†(1๐‘– ฬ‚ + 2๐‘– ฬ‚ โˆ’ 5๐‘˜ ฬ‚) ๐’“ โƒ— = (๐’Š ฬ‚ + 2๐’‹ ฬ‚ + 3๐’Œ ฬ‚) + k (๐’Š ฬ‚ + 2๐’‹ ฬ‚ โˆ’ 5๐’Œ ฬ‚)

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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.