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Class 12
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The equation of the line in vector form passing through the pointΒ  (βˆ’1, 3, 5) and parallel to line (x - 3)/2 = (y - 4)/3, z = 2. is

(a) r = (βˆ’i + 3j + 5k ) + Ξ» (2i + 3j + k )

(b) r = (βˆ’i + 3j + 5k ) + Ξ» (2i + 3j )

(c) r = (2i + 3j βˆ’ 2k ) + Ξ» (βˆ’i + 3j + 5k )

(d) rΒ  = (2iΒ  + 3jΒ  ) + Ξ» (βˆ’iΒ  + 3jΒ  + 5kΒ  )

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


Transcript

Question 10 The equation of the line in vector form passing through the point (βˆ’1, 3, 5) and parallel to line (π‘₯ βˆ’ 3)/2 = (𝑦 βˆ’ 4)/3, z = 2. is (a) π‘Ÿ βƒ— = (βˆ’π‘– Μ‚ + 3𝑗 Μ‚ + 5π‘˜ Μ‚) + πœ† (2𝑖 Μ‚ + 3𝑗 Μ‚ + π‘˜ Μ‚) (b) π‘Ÿ βƒ— = (βˆ’π‘– Μ‚ + 3𝑗 Μ‚ + 5π‘˜ Μ‚) + πœ† (2𝑖 Μ‚ + 3𝑗 Μ‚) (c) π‘Ÿ βƒ— = (2𝑖 Μ‚ + 3𝑗 Μ‚ βˆ’ 2π‘˜ Μ‚) + πœ† (βˆ’π‘– Μ‚ + 3𝑗 Μ‚ + 5π‘˜ Μ‚) (d) π‘Ÿ βƒ— = (2𝑖 Μ‚ + 3𝑗 Μ‚) + πœ† (βˆ’π‘– Μ‚ + 3𝑗 Μ‚ + 5π‘˜ Μ‚) Equation of a line passing through a point with position vector π‘Ž βƒ— and parallel to vector 𝑏 βƒ— is π‘Ÿ βƒ— = π‘Ž βƒ— + πœ†π‘ βƒ— Here, Point is (–1, 3, 5) So, π‘Ž βƒ— = –𝑖 Μ‚ + 3𝑗 Μ‚ + 5π‘˜ Μ‚ And it is parallel to line (π‘₯ βˆ’ 3)/2 = (𝑦 βˆ’ 4)/3, z = 2 This means that z-component of equation is 0 So, line is (π‘₯ βˆ’ 3)/2 = (𝑦 βˆ’ 4)/3 = (𝑧 βˆ’ 2)/0 So, 𝑏 βƒ— = 2𝑖 Μ‚ + 3𝑗 Μ‚ + 0π‘˜ Μ‚ = 2𝑖 Μ‚ + 3𝑗 Μ‚ Thus, Equation of line is 𝒓 βƒ— = (βˆ’π’Š Μ‚ + 3𝒋 Μ‚ + 5π’Œ Μ‚) + 𝝀 (2π’Š Μ‚ + 3𝒋 Μ‚) So, (b) is the correct answer

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.