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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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Misc 3 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

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Davneet Singh

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.