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

Transcript

Ex 11.2, 5 Find the equation of the line in vector and in cartesian form that passes through the point with position vector 2๐‘– ฬ‚ โˆ’ ๐‘— ฬ‚ + 4๐‘˜ ฬ‚ and is in the direction ๐‘– ฬ‚ + 2 ๐‘— ฬ‚ โˆ’ ๐‘˜ ฬ‚ . Equation of a line passing though a point with position vector ๐‘Ž โƒ— and parallel to vector ๐‘ โƒ— is ๐‘Ÿ โƒ— = ๐‘Ž โƒ— + ๐œ† ๐‘ โƒ— Here, ๐‘Ž โƒ— = 2๐‘– ฬ‚ โˆ’ ๐‘— ฬ‚ + 4๐‘˜ ฬ‚ & ๐‘ โƒ— = ๐‘– ฬ‚ + 2๐‘— ฬ‚ โˆ’ ๐‘˜ ฬ‚ So, ๐‘Ÿ โƒ— = (2๐’Š ฬ‚ โˆ’ ๐’‹ ฬ‚ + 4๐’Œ ฬ‚) + ๐œ† (๐’Š ฬ‚ + 2๐’‹ ฬ‚ โˆ’ ๐’Œ ฬ‚) โˆด Equation of line in vector form is (2๐‘– ฬ‚ โˆ’ ๐‘— ฬ‚ + 4๐‘˜ ฬ‚) + ๐œ† (๐‘– ฬ‚ + 2๐‘— ฬ‚ โˆ’ ๐‘˜ ฬ‚) Equation of a line passing though (x1, y1, z1) and parallel to a line having direction ratios a, b, c is (๐‘ฅ โˆ’ ๐‘ฅ1)/๐‘Ž = (๐‘ฆ โˆ’ ๐‘ฆ1)/๐‘ = (๐‘ง โˆ’ ๐‘ง1)/๐‘ Since the line passes through a point with position vector 2๐‘– ฬ‚ โˆ’ ๐‘— ฬ‚ + 4๐‘˜ ฬ‚, โˆด ๐‘ฅ1 = 2, y1 = โˆ’1, z1 = 4 Also, line is in the direction of ๐‘– ฬ‚ + 2๐‘— ฬ‚ โˆ’ ๐‘˜ ฬ‚, Direction ratios : ๐‘Ž = 1, b = 2, c = โˆ’1 Equation of line in Cartesian form is (๐‘ฅ โˆ’ 2)/1 = (๐‘ฆ โˆ’ ( โˆ’1))/2 = (๐‘ง โˆ’ 4)/( โˆ’ 1) (๐’™ โˆ’ ๐Ÿ)/๐Ÿ = (๐’š + ๐Ÿ)/๐Ÿ = (๐’› โˆ’ ๐Ÿ’)/( โˆ’ 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.