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

Transcript

Ex 11.2, 12 Find the values of ๐‘ so that the lines (1 โˆ’ ๐‘ฅ)/3 = (7๐‘ฆ โˆ’ 14)/2๐‘ =(๐‘ง โˆ’ 3)/2 and (7 โˆ’ 7๐‘ฅ)/3๐‘ = (๐‘ฆ โˆ’ 5)/1 = (6 โˆ’ ๐‘ง)/5 are at right angles. 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 at right angles to each other if ๐’‚๐Ÿ ๐’‚๐Ÿ + ๐’ƒ๐Ÿ ๐’ƒ๐Ÿ + ๐’„๐Ÿ ๐’„๐Ÿ = 0 (๐Ÿ โˆ’ ๐’™)/๐Ÿ‘ = (๐Ÿ•๐’š โˆ’ ๐Ÿ๐Ÿ’)/๐Ÿ๐’‘ = (๐’› โˆ’ ๐Ÿ‘)/๐Ÿ ( โˆ’(๐‘ฅ โˆ’ 1))/3 = (7(๐‘ฆ โˆ’ 2))/2๐‘ = (๐‘ง โˆ’ 3)/2 (๐‘ฅ โˆ’ 1)/( โˆ’3) = (๐‘ฆ โˆ’ 2)/(2๐‘/7) = (๐‘ง โˆ’ 3)/2 Comparing with (๐‘ฅ โˆ’ ๐‘ฅ1)/๐‘Ž1 = (๐‘ฆ โˆ’ ๐‘ฆ1)/๐‘1 = (๐‘ง โˆ’ ๐‘ง1)/๐‘1 ๐‘ฅ1 = 1, y1 = 2, z1 = 3 & a1 = โˆ’3, b1 = 2๐‘/7 , c1 = 2 (๐Ÿ• โˆ’ ๐Ÿ•๐’™)/๐Ÿ‘๐’‘ = (๐’š โˆ’ ๐Ÿ“)/๐Ÿ = (๐Ÿ” โˆ’ ๐’›)/๐Ÿ“ ( โˆ’7(๐‘ฅ โˆ’ 1))/3๐‘ = (๐‘ฆ โˆ’ 5)/1 = ( โˆ’ (๐‘ง โˆ’ 6))/5 (๐’™ โˆ’ ๐Ÿ)/( (โˆ’๐Ÿ‘๐’‘)/๐Ÿ•) = (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, x2 = 1, y2 = 5, z2 = 6 & ๐‘Ž2 = ( โˆ’ 3๐‘)/7, b2 = 1, c2 = โˆ’5 Since the lines are perpendicular ๐‘Ž1๐‘Ž2+๐‘1๐‘2+๐‘1๐‘2 = 0 (โˆ’3ร—( โˆ’ 3๐‘)/7) + (2๐‘/7ร—1 ) + (2 ร— โˆ’5) = 0 9๐‘/7 + 2๐‘/7 โˆ’ 10 = 0 11๐‘/7 = 10 p = 10 ร— 7/11 โˆด p = ๐Ÿ•๐ŸŽ/๐Ÿ๐Ÿ

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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.