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Last updated at Feb. 1, 2020 by Teachoo
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Misc 6 If the lines (๐ฅ โ 1)/( โ 3) = (๐ฆ โ 2)/2๐ = (๐ง โ 3)/2 and (๐ฅ โ 1)/3๐ = (๐ฆ โ 1)/1 = (๐ง โ 6)/( โ 5) are perpendicular, find the value of k. Two lines (๐ฅ โ ๐ฅ1)/( ๐1) = (๐ฆ โ๐ฆ1)/๐1 = (๐ง โ ๐ง1)/๐1 and (๐ฅ โ ๐ฅ2)/( ๐2) = (๐ฆ โ ๐ฆ2)/๐2 = (๐ง โ ๐ง2)/๐2 are perpendicular to each other if ๐๐ ๐๐ + ๐๐ ๐๐ + ๐๐ ๐๐ = 0 (๐ โ ๐)/( โ ๐) = (๐ โ ๐)/( ๐๐) = (๐ โ ๐)/( ๐) Comparing with (๐ฅ โ ๐ฅ1)/( ๐1) = (๐ฆโ ๐ฆ1)/( ๐1) = (๐งโ ๐ง1)/( ๐1) x1 = 1, y1 = 2, z1 = 3 & ๐๐ = โ3, b1 = 2k c1 = 2 (๐ โ ๐)/( ๐๐) = (๐ โ ๐)/( ๐) = (๐ โ ๐)/( โ ๐) Comparing with (๐ฅ โ ๐ฅ2)/( ๐2) = (๐ฆ โ ๐ฆ2)/( ๐2) = (๐ง โ ๐ง2)/( ๐2), ๐ฅ2 = 1, y2 = 1, z2 = 6 & ๐๐ = 3k, b2 = 1, c2 = โ5, Since the two lines are perpendicular, ๐1 ๐2 + ๐1 ๐2 + c1๐2 = 0 (โ3 ร 3k) + (2k ร 1) + (2 ร โ 5) = 0 โ 9k + 2k โ 10 = 0 โ 7k = 10 k = (โ๐๐)/๐ Therefore, k = (โ10)/7 Misc 7 Find the vector equation of the line passing through (1, 2, 3) and perpendicular to the plane ๐ โ.(๐ ฬ + 2๐ ฬ โ 5๐ ฬ) + 9 = 0 . The vector equation of a line passing through a point with position vector ๐ โ and parallel to vector ๐ โ is ๐ โ = ๐ โ + ๐๐ โ Given, the line passes (1, 2, 3) So, ๐ โ = 1๐ ฬ + 2๐ ฬ + 3๐ ฬ Finding normal of plane ๐ โ.(๐ ฬ + 2๐ ฬ โ 5๐ ฬ) + 9 = 0 ๐ โ. (๐ ฬ + 2๐ ฬ โ 5๐ ฬ) = โ 9 โ๐ โ. (๐ ฬ + 2๐ ฬ โ 5๐ ฬ) = 9 ๐ โ. (โ1๐ ฬ โ 2๐ ฬ + 5๐ ฬ) = 9 Comparing with ๐ โ. ๐ โ = d, ๐ โ = โ๐ ฬ โ 2๐ ฬ + 5๐ ฬ Since line is perpendicular to plane, the line will be parallel to the normal of the plane โด ๐ โ = ๐ โ = โ1๐ ฬ โ 2๐ ฬ + 5๐ ฬ Hence, ๐ โ = (1๐ ฬ + 2๐ ฬ + 3๐ ฬ) + ๐(โ1๐ ฬ โ 2๐ ฬ + 5๐ ฬ) ๐ โ = (1๐ ฬ + 2๐ ฬ + 3๐ ฬ) โ ๐(1๐ ฬ + 2๐ ฬ โ 5๐ ฬ) ๐ โ = (๐ ฬ + 2๐ ฬ + 3๐ ฬ) + k (๐ ฬ + 2๐ ฬ โ 5๐ ฬ)
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