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

(a) r = (āˆ’i + 3j + 5k ) + Ī» (2i + 3j + k )

(b) r = (āˆ’i + 3j + 5k ) + Ī» (2i + 3j )

(c) r = (2i + 3j āˆ’ 2k ) + Ī» (āˆ’i + 3j + 5k )

(d) rĀ  = (2iĀ  + 3jĀ  ) + Ī» (āˆ’iĀ  + 3jĀ  + 5kĀ  )

The equation of line in vector form passing through point (-1, 3,5)

Question 10 - CBSE Class 12 Sample Paper for 2020 Boards - Part 2

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Question 10 The equation of the line in vector form passing through the point (āˆ’1, 3, 5) and parallel to line (š‘„ āˆ’ 3)/2 = (š‘¦ āˆ’ 4)/3, z = 2. is (a) š‘Ÿ āƒ— = (āˆ’š‘– Ģ‚ + 3š‘— Ģ‚ + 5š‘˜ Ģ‚) + šœ† (2š‘– Ģ‚ + 3š‘— Ģ‚ + š‘˜ Ģ‚) (b) š‘Ÿ āƒ— = (āˆ’š‘– Ģ‚ + 3š‘— Ģ‚ + 5š‘˜ Ģ‚) + šœ† (2š‘– Ģ‚ + 3š‘— Ģ‚) (c) š‘Ÿ āƒ— = (2š‘– Ģ‚ + 3š‘— Ģ‚ āˆ’ 2š‘˜ Ģ‚) + šœ† (āˆ’š‘– Ģ‚ + 3š‘— Ģ‚ + 5š‘˜ Ģ‚) (d) š‘Ÿ āƒ— = (2š‘– Ģ‚ + 3š‘— Ģ‚) + šœ† (āˆ’š‘– Ģ‚ + 3š‘— Ģ‚ + 5š‘˜ Ģ‚) Equation of a line passing through a point with position vector š‘Ž āƒ— and parallel to vector š‘ āƒ— is š‘Ÿ āƒ— = š‘Ž āƒ— + šœ†š‘ āƒ— Here, Point is (–1, 3, 5) So, š‘Ž āƒ— = ā€“š‘– Ģ‚ + 3š‘— Ģ‚ + 5š‘˜ Ģ‚ And it is parallel to line (š‘„ āˆ’ 3)/2 = (š‘¦ āˆ’ 4)/3, z = 2 This means that z-component of equation is 0 So, line is (š‘„ āˆ’ 3)/2 = (š‘¦ āˆ’ 4)/3 = (š‘§ āˆ’ 2)/0 So, š‘ āƒ— = 2š‘– Ģ‚ + 3š‘— Ģ‚ + 0š‘˜ Ģ‚ = 2š‘– Ģ‚ + 3š‘— Ģ‚ Thus, Equation of line is š’“ āƒ— = (āˆ’š’Š Ģ‚ + 3š’‹ Ģ‚ + 5š’Œ Ģ‚) + š€ (2š’Š Ģ‚ + 3š’‹ Ģ‚) So, (b) is the correct answer

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