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Misc 6 - If lines are perpendicular find value of k. x-1/-3 = y-2/2k

Misc 6 - Chapter 11 Class 12 Three Dimensional Geometry - Part 2

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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.