Misc 10 - Find coordinates of point where line through (5, 1, 6)

Misc 10 - Chapter 11 Class 12 Three Dimensional Geometry - Part 2
Misc 10 - Chapter 11 Class 12 Three Dimensional Geometry - Part 3

Misc 10 - Chapter 11 Class 12 Three Dimensional Geometry - Part 4 Misc 10 - Chapter 11 Class 12 Three Dimensional Geometry - Part 5 Misc 10 - Chapter 11 Class 12 Three Dimensional Geometry - Part 6

 

  1. Chapter 11 Class 12 Three Dimensional Geometry
  2. Serial order wise

Transcript

Misc 10 (Method 1) Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the YZ-plane.The equation of a line passing through two points with position vectors ๐‘Ž โƒ— & ๐‘ โƒ— is ๐’“ โƒ— = ๐’‚ โƒ— + ๐œ†(๐’ƒ โƒ— โˆ’ ๐’‚ โƒ—) Given, the line passes through (๐‘ โƒ— โˆ’ ๐‘Ž โƒ—) = (3๐‘– ฬ‚ + 4๐‘— ฬ‚ + 1๐‘˜ ฬ‚) โˆ’ (5๐‘– ฬ‚ + 1๐‘— ฬ‚ + 6๐‘˜ ฬ‚) = (3 โˆ’5)๐‘– ฬ‚ + (4 โˆ’ 1)๐‘— ฬ‚ + (1 โˆ’ 6)๐‘˜ ฬ‚ = โˆ’2๐‘– ฬ‚ + 3๐‘— ฬ‚ โˆ’ 5๐‘˜ ฬ‚ The equation of a line passing through two points with position vectors ๐‘Ž โƒ— & ๐‘ โƒ— is ๐’“ โƒ— = ๐’‚ โƒ— + ๐œ†(๐’ƒ โƒ— โˆ’ ๐’‚ โƒ—) Given, the line passes through (๐‘ โƒ— โˆ’ ๐‘Ž โƒ—) = (3๐‘– ฬ‚ + 4๐‘— ฬ‚ + 1๐‘˜ ฬ‚) โˆ’ (5๐‘– ฬ‚ + 1๐‘— ฬ‚ + 6๐‘˜ ฬ‚) = (3 โˆ’5)๐‘– ฬ‚ + (4 โˆ’ 1)๐‘— ฬ‚ + (1 โˆ’ 6)๐‘˜ ฬ‚ = โˆ’2๐‘– ฬ‚ + 3๐‘— ฬ‚ โˆ’ 5๐‘˜ ฬ‚ B(3, 4, 1) ๐‘ โƒ— = 3๐‘– ฬ‚ + 4๐‘— ฬ‚ + 1๐‘˜ ฬ‚ โˆด ๐’“ โƒ— = (5๐’Š ฬ‚ + ๐’‹ ฬ‚ + 6๐’Œ ฬ‚) + ๐œ† (โˆ’2๐’Š ฬ‚ + 3๐’‹ ฬ‚ โˆ’ 5๐’Œ ฬ‚) Let the coordinates of the point where the line crosses the YZ plane be (0, y, z) So, ๐’“ โƒ— = 0๐’Š ฬ‚ + y๐’‹ ฬ‚ + z๐’Œ ฬ‚ Since point lies in line, it will satisfy its equation, Putting (2) in (1) 0๐‘– ฬ‚ + y๐‘— ฬ‚ + z๐‘˜ ฬ‚ = 5๐‘– ฬ‚ + ๐‘— ฬ‚ + 6๐‘˜ ฬ‚ โˆ’2๐œ†๐‘– ฬ‚ + 3๐œ†๐‘— ฬ‚ โˆ’ 5๐œ†๐‘˜ ฬ‚ 0๐‘– ฬ‚ + y๐‘— ฬ‚ + z๐‘˜ ฬ‚ = (5 โˆ’2๐œ†)๐‘– ฬ‚ + (1 + 3๐œ†)๐‘— ฬ‚ + (6 โˆ’ 5๐œ†)๐‘˜ ฬ‚ Two vectors are equal if their corresponding components are equal So, Solving 0 = 5 โˆ’ 2๐œ† 5 = 2๐œ† โˆด ๐œ† = ๐Ÿ“/๐Ÿ Now, y = 1 + 3๐œ† = 1 + 3 ร— 5/2 = 1 + 15/2 = 17/2 & z = 6 โˆ’ 5๐œ† = 6 โˆ’ 5 ร— 5/2 = 6 โˆ’ 25/2 = (โˆ’13)/2 Therefore, the coordinates of the required point is (๐ŸŽ,๐Ÿ๐Ÿ•/๐Ÿ, (โˆ’๐Ÿ๐Ÿ‘)/๐Ÿ). Misc 10 (Method 2) Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the YZ-plane.The equation of a line passing through two points A(๐‘ฅ_1, ๐‘ฆ_1, ๐‘ง_1) and B(๐‘ฅ_2, ๐‘ฆ_2, ๐‘ง_2) is (๐’™ โˆ’ ๐’™_๐Ÿ)/(๐’™_๐Ÿ โˆ’ ๐’™_๐Ÿ ) = (๐’š โˆ’ ๐’š_๐Ÿ)/(๐’š_๐Ÿ โˆ’ ๐’š_๐Ÿ ) = (๐’› โˆ’ ๐’›_๐Ÿ)/(๐’›_๐Ÿ โˆ’ ๐’›_๐Ÿ ) Given the line passes through the points So, the equation of line is (๐‘ฅ โˆ’ 5)/(3 โˆ’ 5) = (๐‘ฆ โˆ’ 1)/(4 โˆ’ 1) = (๐‘ง โˆ’ 6)/(1 โˆ’ 6) A (5, 1, 6) โˆด ๐‘ฅ_1= 5, ๐‘ฆ_1= 1, ๐‘ง_1= 6 B(3, 4, 1) โˆด ๐‘ฅ_2= 3, ๐‘ฆ_2= 4, ๐‘ง_2= 1 (๐’™ โˆ’ ๐Ÿ“)/(โˆ’๐Ÿ) = (๐’š โˆ’ ๐Ÿ)/๐Ÿ‘ = (๐’› โˆ’ ๐Ÿ”)/(โˆ’๐Ÿ“) = k So, Since the line crosses the YZ plane at (0, y, z) x = 0 โˆ’2k + 5 = 0 2k = 5 k = ๐Ÿ“/๐Ÿ x = โ€“2k + 5 So, x = โˆ’2k + 5 = โˆ’2 ร— 5/2 + 5 = โˆ’ 10/2 + 5 = 0 y = 3k + 1 = 3 ร— 5/2 + 1 = 15/2 + 1 = 17/2 & z = โˆ’5k + 6 = โˆ’5 ร— 5/2 + 6 = (โˆ’25)/2 + 6 = (โˆ’13)/2 Therefore, the coordinates of the required point are (๐ŸŽ,๐Ÿ๐Ÿ•/๐Ÿ,(โˆ’๐Ÿ๐Ÿ‘)/๐Ÿ).

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.