Last updated at March 11, 2017 by Teachoo

Transcript

Example 12 Find the distance between the lines 𝑙1 and 𝑙2 given by 𝑟 = 𝑖 + 2 𝑗 – 4 𝑘 + 𝜆(2 𝒊 + 3 𝒋 + 6 𝒌 ) and 𝑟 = 3 𝑖 + 3 𝑗 − 5 𝑘 + μ (2 𝒊 + 3 𝒋 + 6 𝒌) Distance between two parallel lines with vector equations 𝑟 = 𝑎1 + 𝜆 𝑏 and 𝑟 = 𝑎2 + μ 𝑏 is 𝑏 × ( 𝑎2 − 𝑎1) 𝑏 Now, ( 𝒂𝟐 − 𝒂𝟏) = (3 𝑖 + 3 𝑗 − 5 𝑘) − (1 𝑖 + 2 𝑗 − 4 𝑘) = (3 − 1) 𝑖 + (3 − 2) 𝑗 + ( − 5 + 4) 𝑘 = 2 𝒊 + 1 𝒋 − 1 𝒌 Magnitude of 𝑏 = 22 + 32 + 62 𝒃 = 4+9+36 = 49 = 7 Also, 𝒃 × ( 𝒂𝟐 − 𝒂𝟏) = 𝑖 𝑗 𝑘23621−1 = 𝑖 3×−1−(1×6) − 𝑗 2×−1−(2×6) + 𝑘 2×1−(2×3) = 𝑖 −3−6 − 𝑗 −2−12 + 𝑘 2−6 = 𝑖 (–9) − 𝑗 (–14) + 𝑘(−4) = −𝟗 𝒊 + 14 𝒋 − 4 𝒌 Now, 𝒃 × ( 𝒂𝟐 − 𝒂𝟏) = −92+ 142+ −42 = 81+196+16 = 𝟐𝟗𝟑 So, distance = 𝑏 × ( 𝑎2 − 𝑎1) 𝑏 = 2937 = 𝟐𝟗𝟑𝟕 Therefore, the distance between the given two parallel lines is 2937.

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Chapter 11 Class 12 Three Dimensional Geometry

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 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.