The equation of the line in vector form passing through the point  (−1, 3, 5) and parallel to line (x - 3)/2 = (y - 4)/3, z = 2. is

(a) r = (−i + 3j + 5k ) + λ (2i + 3j + k )

(b) r = (−i + 3j + 5k ) + λ (2i + 3j )

(c) r = (2i + 3j − 2k ) + λ (−i + 3j + 5k )

(d) r  = (2i  + 3j  ) + λ (−i  + 3j  + 5k  )

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  1. Class 12
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Transcript

Question 10 The equation of the line in vector form passing through the point (โˆ’1, 3, 5) and parallel to line (๐‘ฅ โˆ’ 3)/2 = (๐‘ฆ โˆ’ 4)/3, z = 2. is (a) ๐‘Ÿ โƒ— = (โˆ’๐‘– ฬ‚ + 3๐‘— ฬ‚ + 5๐‘˜ ฬ‚) + ๐œ† (2๐‘– ฬ‚ + 3๐‘— ฬ‚ + ๐‘˜ ฬ‚) (b) ๐‘Ÿ โƒ— = (โˆ’๐‘– ฬ‚ + 3๐‘— ฬ‚ + 5๐‘˜ ฬ‚) + ๐œ† (2๐‘– ฬ‚ + 3๐‘— ฬ‚) (c) ๐‘Ÿ โƒ— = (2๐‘– ฬ‚ + 3๐‘— ฬ‚ โˆ’ 2๐‘˜ ฬ‚) + ๐œ† (โˆ’๐‘– ฬ‚ + 3๐‘— ฬ‚ + 5๐‘˜ ฬ‚) (d) ๐‘Ÿ โƒ— = (2๐‘– ฬ‚ + 3๐‘— ฬ‚) + ๐œ† (โˆ’๐‘– ฬ‚ + 3๐‘— ฬ‚ + 5๐‘˜ ฬ‚) Equation of a line passing through 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, Point is (โ€“1, 3, 5) So, ๐‘Ž โƒ— = โ€“๐‘– ฬ‚ + 3๐‘— ฬ‚ + 5๐‘˜ ฬ‚ And it is parallel to line (๐‘ฅ โˆ’ 3)/2 = (๐‘ฆ โˆ’ 4)/3, z = 2 This means that z-component of equation is 0 So, line is (๐‘ฅ โˆ’ 3)/2 = (๐‘ฆ โˆ’ 4)/3 = (๐‘ง โˆ’ 2)/0 So, ๐‘ โƒ— = 2๐‘– ฬ‚ + 3๐‘— ฬ‚ + 0๐‘˜ ฬ‚ = 2๐‘– ฬ‚ + 3๐‘— ฬ‚ Thus, Equation of line is ๐’“ โƒ— = (โˆ’๐’Š ฬ‚ + 3๐’‹ ฬ‚ + 5๐’Œ ฬ‚) + ๐€ (2๐’Š ฬ‚ + 3๐’‹ ฬ‚) So, (b) is the correct answer

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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.