Ex 11.2, 7 - Cartesian equation is  x-5/3 = y+4/7 = z-6/2 - Ex 11.2

part 2 - Ex 11.2, 7 - Ex 11.2 - Serial order wise - Chapter 11 Class 12 Three Dimensional Geometry

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Transcript

Ex 11.2, 7 The Cartesian equation of a line is (๐‘ฅ โˆ’ 5)/3 = (๐‘ฆ + 4)/7 = (๐‘ง โˆ’ 6)/2. Write its vector form.Cartesian equation : (๐‘ฅ โˆ’ 5)/3 = (๐‘ฆ + 4)/7 = (๐‘ง โˆ’ 6)/2 (๐’™ โˆ’ ๐Ÿ“)/๐Ÿ‘ = (๐’š โˆ’ (โˆ’๐Ÿ’))/๐Ÿ• = (๐’› โˆ’ ๐Ÿ”)/๐Ÿ Equation of a line in Cartesian form is given by (๐‘ฅ โˆ’ ๐‘ฅ1)/๐‘Ž = (๐‘ฆ โˆ’ ๐‘ฆ1)/๐‘ = (๐‘ง โˆ’ ๐‘ง1)/๐‘ Comparing (1) and (2), ๐’™1 = 5, ๐’š1 = โˆ’4, ๐’›1 = 6 ๐’‚ = 3, b = 7, c = 2 Equation of line in vector form is ๐’“ โƒ— = ๐’‚ โƒ— + ๐œ†๐’ƒ โƒ— Where ๐’‚ โƒ— = ๐‘ฅ1๐‘– ฬ‚ + y1๐‘— ฬ‚ + z1๐‘˜ ฬ‚ = 5๐’Š ฬ‚ โˆ’ 4๐’‹ ฬ‚ + 6๐’Œ ฬ‚ & ๐’ƒ โƒ— = ๐‘Ž๐‘– ฬ‚ + b๐‘— ฬ‚ + c๐‘˜ ฬ‚ = 3๐’Š ฬ‚ + 7๐’‹ ฬ‚ + 2๐’Œ ฬ‚ Now, ๐‘Ÿ โƒ— = (5๐’Š ฬ‚ โˆ’ 4๐’‹ ฬ‚ + 6๐’Œ ฬ‚) + ๐œ† (3๐’Š ฬ‚ + 7๐’‹ ฬ‚ + 2๐’Œ ฬ‚)

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