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Last updated at Feb. 1, 2020 by Teachoo
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Ex 11.2, 9 Find the vector and the Cartesian equations of the line that passes through the points (3, โ 2, โ 5), (3, โ 2, 6).Vector Equation Vector equation of a line passing through two points with position vectors ๐ โ and ๐ โ is ๐ โ = ๐ โ + ๐ (๐ โ โ ๐ โ) Given, the two points are So, ๐ โ = (3๐ ฬ โ 2๐ ฬ โ 5๐ ฬ) + ๐ ["(3" ๐ ฬโ"2" ๐ ฬ+"6" ๐ ฬ")" โ"(3" ๐ ฬโ"2" ๐ ฬ โ"5" ๐ ฬ")" ] = 3๐ ฬ โ 2๐ ฬ โ 5๐ ฬ + ๐ ["(3" โ3")" ๐ ฬโ"(2" โ(โ2))๐ ฬ+(6โ(โ5))๐ ฬ)] A (3, โ 2, โ 5) ๐ โ = 3๐ ฬ โ 2๐ ฬ โ 5๐ ฬ B (3, โ 2, 6) ๐ โ = 3๐ ฬ โ 2๐ ฬ + 6๐ ฬ = 3๐ ฬ โ 2๐ ฬ โ 5๐ ฬ + ๐ [0๐ ฬ + 0๐ ฬ + 11๐ ฬ] = 3๐ ฬ โ 2๐ ฬ โ 5๐ ฬ + ๐ (11๐ ฬ) Therefore, the vector equation is ๐ โ = 3๐ ฬ โ 2๐ ฬ โ 5๐ ฬ + ๐ (11๐ ฬ) Cartesian equation Cartesian equation of a line passing through two points A(x1, y1, z1) and B (x2, y2, z2) is (๐ฅ โ ๐ฅ1)/(๐ฅ2 โ ๐ฅ_1 ) = (๐ฆ โ ๐ฆ1)/(๐ฆ2 โ ๐ฆ1) = (๐ง โ ๐ง1)/(๐ง2 โ ๐ง1) Since the line passes through A (3, โ2, โ5) x1 = 3, y1 = โ2, z1 = โ 5 And also passes through B (3, โ2, 6) x2 = 3, y2 = โ2, z2 = 6 Equation of line is (๐ฅ โ 3)/(3 โ 3) = (๐ฆ โ (โ2))/( โ2 โ (โ2)) = (๐ง โ (โ5))/(6 โ (โ5)) (๐ โ ๐)/๐ = (๐ + ๐)/๐ = (๐ + ๐)/๐๐
Ex 11.2
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Ex 11.2, 10 Important Not in Syllabus - CBSE Exams 2021
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