Last updated at Aug. 23, 2021 by Teachoo

Transcript

Example 10 Find the angle between the pair of lines (๐ฅ + 3)/3 = (๐ฆ โ 1)/5 = (๐ง + 3)/4 and (๐ฅ + 1)/1 = (๐ฆ โ 4)/1 = (๐ง โ 5)/2Angle 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 โ ๐)/๐ = (๐ + ๐)/๐ (๐ฅ โ (โ3))/3 = (๐ฆ โ 1)/5 = (๐ง โ (โ3))/4 Comparing with (๐ฅ โ ๐ฅ1)/๐1 = (๐ฆ โ ๐ฆ1)/๐1 = (๐ง โ ๐ง1)/๐1 x1 = โ3, y1 = 1, z1 = โ3 & ๐1 = 3, b1 = 5, c1 = 4 (๐ + ๐)/๐ = (๐ โ ๐)/๐ = (๐ โ ๐)/๐ (๐ฅ โ (โ1))/1 = (๐ฆ โ 4)/1 = (๐ง โ 5)/2 Comparing with (๐ฅ โ ๐ฅ2)/๐2 = (๐ฆ โ ๐ฆ2)/๐2 = (๐ง โ ๐ง2)/๐2 ๐ฅ2 = โ1, y2 = 4, z2 = 5 & ๐2 = 1, ๐2 = 1, ๐2 = 2 Now, cos ฮธ = |(๐_1 ๐_2 + ๐_1 ๐_2 +ใ ๐ใ_1 ๐_2)/(โ(ใ๐_1ใ^2 + ใ๐_1ใ^2+ ใ๐_1ใ^2 ) โ(ใ๐_2ใ^2 +ใใ ๐ใ_2ใ^2+ ใ๐_2ใ^2 ))| = |((3 ร 1) + (5 ร 1) + (4 ร 2))/(โ(3^2 + 5^2 + 4^2 ) ร โ(1^2 + 1^2 + 2^2 ))| = |(3 + 5 + 8)/(โ(9 + 25 + 16) โ(1 + 1 + 4))| = |16/(โ50 โ6)| = |16/(5โ2 ร โ2 โ3)| = |16/(5 ร 2 ร โ3)| = 8/(5 โ3) = 8/(5 โ3) ร โ3/โ3 = (8โ3)/(15 ) So, cos ฮธ = (8โ3)/(15 ) โด ฮธ = cos-1((๐โ๐)/(๐๐ )) Therefore, the angle between the given pair of line is cosโ1 ((8โ3)/(15 ))

Examples

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Example, 9 Deleted for CBSE Board 2022 Exams

Example 10 Important Deleted for CBSE Board 2022 Exams You are here

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Example 22 Deleted for CBSE Board 2022 Exams

Example 23 Important Deleted for CBSE Board 2022 Exams

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Chapter 11 Class 12 Three Dimensional Geometry (Term 2)

Serial order wise

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.