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CBSE Class 12 Sample Paper for 2020 Boards

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Question 10 - CBSE Class 12 Sample Paper for 2020 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Last updated at March 23, 2023 by

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

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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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