

Subscribe to our Youtube Channel - https://you.tube/teachoo
Last updated at Feb. 1, 2020 by Teachoo
Transcript
Angle between the pair of 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 is given by cos ฮธ = |(๐_๐ ๐_๐ + ๐_๐ ๐_๐ +ใ ๐ใ_๐ ๐_๐)/(โ(ใ๐_๐ใ^๐ + ใ๐_๐ใ^๐+ ใ๐_๐ใ^๐ ) โ(ใ๐_๐ใ^๐ +ใใ ๐ใ_๐ใ^๐+ ใ๐_๐ใ^๐ ))| Angle between the pair of 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 is given by cos ฮธ = |(๐_๐ ๐_๐ + ๐_๐ ๐_๐ +ใ ๐ใ_๐ ๐_๐)/(โ(ใ๐_๐ใ^๐ + ใ๐_๐ใ^๐+ ใ๐_๐ใ^๐ ) โ(ใ๐_๐ใ^๐ +ใใ ๐ใ_๐ใ^๐+ ใ๐_๐ใ^๐ ))| (๐ โ ๐)/๐ = (๐ โ ๐)/๐ = (๐ + ๐)/( โ ๐) (๐ฅ โ 2)/2 = (๐ฆ โ 1)/5 = (๐ง โ (โ3))/( โ 3) Comparing with (๐ฅ โ ๐ฅ1)/๐1 = (๐ฆ โ ๐ฆ1)/๐1 = (๐ง โ ๐ง1)/๐1 x1 = 2, y1 = 1, z1 = โ3 & ๐1 = 2, b1 = 5, c1 = โ3 (๐ + ๐)/( โ ๐) = (๐ โ ๐)/๐ = (๐ โ ๐)/๐ (๐ฅ โ (โ 2))/( โ 1) = (๐ฆ โ 4)/8 = (๐ง โ 5)/4 Comparing with (๐ฅ โ ๐ฅ2)/๐2 = (๐ฆ โ ๐ฆ2)/๐2 = (๐ง โ ๐ง2)/๐2 ๐ฅ2 = โ 2, y2 = 4, z2 = 5 & ๐2 = โ1, ๐2 = 8, ๐2 = 4 Now, cos ฮธ = |(๐_๐ ๐_๐ + ๐_๐ ๐_๐ +ใ ๐ใ_๐ ๐_๐)/(โ(ใ๐_๐ใ^๐ + ใ๐_๐ใ^๐+ ใ๐_๐ใ^๐ ) โ(ใ๐_๐ใ^๐ +ใใ ๐ใ_๐ใ^๐+ ใ๐_๐ใ^๐ ))| = |((2 ร โ1) + (5 ร 8) + ( โ 3 ร 4) )/(โ(2^2 + 5^2 + ใ(โ3)ใ^2 ) โ(ใ(โ1)ใ^2 + 8^2 + 4^2 ))| = |( โ2 + 40 + (โ12) )/(โ(4 + 25 + 9) โ(1 + 64 + 16))| = |26/(โ38 โ81)| = |26/(โ38 ร 9)| = 26/(9โ38 ) So, cos ฮธ = 26/(9โ38 ) โด ฮธ = cosโ1 (๐๐/(๐โ๐๐ )) Therefore, the angle between the given lines is cos-1 (26/(9โ38 )). = |( โ2 + 40 + (โ12) )/(โ(4 + 25 + 9) โ(1 + 64 + 16))| = |26/(โ38 โ81)| = |26/(โ38 ร 9)| = 26/(9โ38 ) So, cos ฮธ = 26/(9โ38 ) โด ฮธ = cosโ1 (๐๐/(๐โ๐๐ )) Therefore, the angle between the given lines is cos-1 (26/(9โ38 )). Angle between the pair of 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 is given by cos ฮธ = |(๐_๐ ๐_๐ + ๐_๐ ๐_๐ +ใ ๐ใ_๐ ๐_๐)/(โ(ใ๐_๐ใ^๐ + ใ๐_๐ใ^๐+ ใ๐_๐ใ^๐ ) โ(ใ๐_๐ใ^๐ +ใใ ๐ใ_๐ใ^๐+ ใ๐_๐ใ^๐ ))| ๐/๐ = ๐/๐ = ๐/๐ (๐ฅ โ 0)/2 = (๐ฆ โ 0)/2 = (๐ง โ 0)/1 Comparing with (๐ฅ โ ๐ฅ1)/๐1 = (๐ฆ โ ๐ฆ1)/๐1 = (๐ง โ ๐ง1)/๐1 x1 = 0, y1 = 0, z1 = 0 & ๐1 = 2, b1 = 2, c1 = 1 (๐ โ ๐)/๐ = (๐ โ ๐)/๐ = (๐ โ ๐)/๐ Comparing with (๐ฅ โ ๐ฅ2)/๐2 = (๐ฆ โ ๐ฆ2)/๐2 = (๐ง โ ๐ง2)/๐2 ๐ฅ2 = 5, y2 = 2, z2 = 5 & ๐2 = 4, ๐2 = 1, ๐2 = 8 Now, cos ฮธ = |(๐_๐ ๐_๐ + ๐_๐ ๐_๐ +ใ ๐ใ_๐ ๐_๐)/(โ(ใ๐_๐ใ^๐ + ใ๐_๐ใ^๐+ ใ๐_๐ใ^๐ ) โ(ใ๐_๐ใ^๐ +ใใ ๐ใ_๐ใ^๐+ ใ๐_๐ใ^๐ ))| = |((2 ร 4) + (2 ร 1) + (1 ร 8) )/(โ(22 + 22 + 12) ร โ(42 + 12 + 82))| = |(8 + 2 + 8 )/(โ(4 + 4 + 1) โ(16 + 1 + 64))| = |18/(โ9 ร โ81)| = 18/(3 ร 9) = 2/3 So, cos ฮธ = 2/3 โด ฮธ = cos-1 (๐/๐) Therefore, the angle between the given lines is cosโ1(๐/๐).
Ex 11.2
Ex 11.2, 2
Ex 11.2, 3 Important
Ex 11.2, 4
Ex 11.2, 5 Important
Ex 11.2, 6
Ex 11.2, 7 Important
Ex 11.2, 8
Ex 11.2, 9 Important
Ex 11.2, 10 Important Not in Syllabus - CBSE Exams 2021
Ex 11.2, 11 Important Not in Syllabus - CBSE Exams 2021 You are here
Ex 11.2, 12 Important
Ex 11.2, 13
Ex 11.2, 14 Important
Ex 11.2, 15 Important
Ex 11.2, 16 Important
Ex 11.2, 17 Important
About the Author