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Last updated at Feb. 1, 2020 by Teachoo
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Ex 11.2, 12 Find the values of ๐ so that the lines (1 โ ๐ฅ)/3 = (7๐ฆ โ 14)/2๐ =(๐ง โ 3)/2 and (7 โ 7๐ฅ)/3๐ = (๐ฆ โ 5)/1 = (6 โ ๐ง)/5 are at right angles. Two lines (๐ฅ โ ๐ฅ1)/๐1 = (๐ฆ โ ๐ฆ1)/๐1 = (๐ง โ ๐ง1)/๐1 and (๐ฅ โ ๐ฅ2)/๐2 = (๐ฆ โ ๐ฆ2)/๐2 = (๐ง โ ๐ง2)/๐2 are at right angles to each other if ๐๐ ๐๐ + ๐๐ ๐๐ + ๐๐ ๐๐ = 0 (๐ โ ๐)/๐ = (๐๐ โ ๐๐)/๐๐ = (๐ โ ๐)/๐ ( โ(๐ฅ โ 1))/3 = (7(๐ฆ โ 2))/2๐ = (๐ง โ 3)/2 (๐ฅ โ 1)/( โ3) = (๐ฆ โ 2)/(2๐/7) = (๐ง โ 3)/2 Comparing with (๐ฅ โ ๐ฅ1)/๐1 = (๐ฆ โ ๐ฆ1)/๐1 = (๐ง โ ๐ง1)/๐1 ๐ฅ1 = 1, y1 = 2, z1 = 3 & a1 = โ3, b1 = 2๐/7 , c1 = 2 (๐ โ ๐๐)/๐๐ = (๐ โ ๐)/๐ = (๐ โ ๐)/๐ ( โ7(๐ฅ โ 1))/3๐ = (๐ฆ โ 5)/1 = ( โ (๐ง โ 6))/5 (๐ โ ๐)/( (โ๐๐)/๐) = (๐ โ ๐)/๐ = (๐ โ ๐)/( โ๐) Comparing with (๐ฅ โ ๐ฅ2)/๐2 = (๐ฆ โ ๐ฆ2)/๐2 = (๐ง โ ๐ง2)/๐2, x2 = 1, y2 = 5, z2 = 6 & ๐2 = ( โ 3๐)/7, b2 = 1, c2 = โ5 Since the lines are perpendicular ๐1๐2+๐1๐2+๐1๐2 = 0 (โ3ร( โ 3๐)/7) + (2๐/7ร1 ) + (2 ร โ5) = 0 9๐/7 + 2๐/7 โ 10 = 0 11๐/7 = 10 p = 10 ร 7/11 โด p = ๐๐/๐๐
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Ex 11.2, 10 Important Not in Syllabus - CBSE Exams 2021
Ex 11.2, 11 Important Not in Syllabus - CBSE Exams 2021
Ex 11.2, 12 Important You are here
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