Misc 11 - Find coordinates of point where line through (5, 1, 6) and

Misc 11 - Chapter 11 Class 12 Three Dimensional Geometry - Part 2
Misc 11 - Chapter 11 Class 12 Three Dimensional Geometry - Part 3 Misc 11 - Chapter 11 Class 12 Three Dimensional Geometry - Part 4 Misc 11 - Chapter 11 Class 12 Three Dimensional Geometry - Part 5 Misc 11 - Chapter 11 Class 12 Three Dimensional Geometry - Part 6

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Question 7 (Method 1) Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the ZX-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‚ A (5, 1, 6) ๐‘Ž โƒ— = 5๐‘– ฬ‚ + 1๐‘— ฬ‚ + 6๐‘˜ ฬ‚ B(3, 4, 1) ๐‘ โƒ— = 3๐‘– ฬ‚ + 4๐‘— ฬ‚ + 1๐‘˜ ฬ‚ = โˆ’2๐‘– ฬ‚ + 3๐‘— ฬ‚ โˆ’ 5๐‘˜ 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 ZX plane be (x, 0, z) So, ๐’“ โƒ— = x๐’Š ฬ‚ + 0๐’‹ ฬ‚ + z๐’Œ ฬ‚ Since point lies in line, it will satisfy its equation, Putting (2) in (1) x๐‘– ฬ‚ + 0๐‘— ฬ‚ + z๐‘˜ ฬ‚ = 5๐‘– ฬ‚ + ๐‘— ฬ‚ + 6๐‘˜ ฬ‚ โˆ’2๐œ†๐‘– ฬ‚ + 3๐œ†๐‘— ฬ‚ โˆ’ 5๐œ†๐‘˜ ฬ‚ x๐‘– ฬ‚ + 0๐‘— 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 = 1 + 3๐œ† 3๐œ† = โˆ’1 โˆด ๐œ† = (โˆ’๐Ÿ)/๐Ÿ‘ Now, x = 5 โˆ’ 2๐œ† = 5 โˆ’ 2 ร— (โˆ’1)/3 = 5 + 2/3 = 17/13 z = 6 โˆ’ 5๐œ† = 6 โˆ’ 5 ร— (โˆ’1)/3 = 6 + 5/3 = 23/3 Therefore, the coordinate of the required point are (๐Ÿ๐Ÿ•/๐Ÿ‘,๐ŸŽ,๐Ÿ๐Ÿ‘/๐Ÿ‘) Question 7 (Method 2) Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the ZX-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 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 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) (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 ZX plane at (x, 0, z), y = 0 3k + 1 = 0 3k = โˆ’1 โˆด k = (โˆ’๐Ÿ)/๐Ÿ‘ So, x = โ€“2k + 5 = โˆ’2 ร— (โˆ’1)/3 + 5 = 2/3 + 5 = 17/3 y = 0 & z = โˆ’5k + 6 = โˆ’5 ร— (โˆ’1)/3 + 6 = 5/3 + 6 = 23/3 therefore, the coordinate of the required point are (๐Ÿ๐Ÿ•/๐Ÿ‘,๐ŸŽ,๐Ÿ๐Ÿ‘/๐Ÿ‘)

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