Last updated at Feb. 1, 2020 by Teachoo
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Misc 11 (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 โ & ๐ โ is ๐ โ = ๐ โ + ๐(๐ โ โ ๐ โ) Given, the line passes through (๐ โ โ ๐ โ) = (3๐ ฬ + 4๐ ฬ + 1๐ ฬ) โ (5๐ ฬ + 1๐ ฬ + 6๐ ฬ) = (3 โ5)๐ ฬ + (4 โ 1)๐ ฬ + (1 โ 6)๐ ฬ A (5, 1, 6) ๐ โ = 5๐ ฬ + 1๐ ฬ + 6๐ ฬ B(3, 4, 1) ๐ โ = 3๐ ฬ + 4๐ ฬ + 1๐ ฬ = โ2๐ ฬ + 3๐ ฬ โ 5๐ ฬ โด ๐ โ = (5๐ ฬ + ๐ ฬ + 6๐ ฬ) + ๐ (โ2๐ ฬ + 3๐ ฬ โ 5๐ ฬ) Let the coordinates of the point where the line crosses the ZX plane be (x, 0, z) So, ๐ โ = x๐ ฬ + 0๐ ฬ + z๐ 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 โ2๐๐ ฬ + 3๐๐ ฬ โ 5๐๐ ฬ x๐ ฬ + 0๐ ฬ + z๐ ฬ = (5 โ2๐)๐ ฬ + (1 + 3๐)๐ ฬ + (6 โ 5๐)๐ ฬ Two vectors are equal if their corresponding components are equal So, Solving 0 = 1 + 3๐ 3๐ = โ1 โด ๐ = (โ๐)/๐ Now, x = 5 โ 2๐ = 5 โ 2 ร (โ1)/3 = 5 + 2/3 = 17/13 z = 6 โ 5๐ = 6 โ 5 ร (โ1)/3 = 6 + 5/3 = 23/3 Therefore, the coordinate of the required point are (๐๐/๐,๐,๐๐/๐) Misc 11 (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(๐ฅ_1, ๐ฆ_1, ๐ง_1) and B(๐ฅ_2, ๐ฆ_2, ๐ง_2) is (๐ โ ๐_๐)/(๐_๐ โ ๐_๐ ) = (๐ โ ๐_๐)/(๐_๐ โ ๐_๐ ) = (๐ โ ๐_๐)/(๐_๐ โ ๐_๐ ) Given the line passes through the points A (5, 1, 6) โด ๐ฅ_1= 5, ๐ฆ_1= 1, ๐ง_1= 6 B(3, 4, 1) โด ๐ฅ_2= 3, ๐ฆ_2= 4, ๐ง_2= 1 So, the equation of line is (๐ฅ โ 5)/(3 โ 5) = (๐ฆ โ 1)/(4 โ 1) = (๐ง โ 6)/(1 โ 6) (๐ โ ๐)/(โ๐) = (๐ โ ๐)/๐ = (๐ โ ๐)/(โ๐) = k So, Since the line crosses the ZX plane at (x, 0, z), y = 0 3k + 1 = 0 3k = โ1 โด k = (โ๐)/๐ So, x = โ2k + 5 = โ2 ร (โ1)/3 + 5 = 2/3 + 5 = 17/3 y = 0 & z = โ5k + 6 = โ5 ร (โ1)/3 + 6 = 5/3 + 6 = 23/3 therefore, the coordinate of the required point are (๐๐/๐,๐,๐๐/๐)
Miscellaneous
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