Last updated at March 11, 2017 by Teachoo

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Ex 11.2, 14 Find the shortest distance between the lines 𝑟 = ( 𝑖 + 2 𝑗 + 𝑘) + 𝜆 ( 𝑖 − 𝑗 + 𝑘) and 𝑟 = (2 𝑖 − 𝑗 − 𝑘) + 𝜇 (2 𝑖 + 𝑗 + 2 𝑘) Shortest distance between the lines with vector equations 𝑟 = 𝑎1 + 𝜆 𝑏1and 𝑟 = 𝑎2 + 𝜇 𝑏2 is 𝒃𝟏 × 𝒃𝟐 . 𝒂𝟐 − 𝒂𝟏 𝒃𝟏 × 𝒃𝟐 Given, Now, 𝒂𝟐 − 𝒂𝟏 = (2 𝑖 − 1 𝑗 − 1 𝑘) − (1 𝑖 + 2 𝑗 + 1 𝑘) = (2 − 1) 𝑖 + (−1− 2) 𝑗 + (−1 − 1) 𝑘 = 1 𝒊 − 3 𝒋 − 2 𝒌 𝒃𝟏 × 𝒃𝟐 = 𝑖 𝑗 𝑘1 − 11212 = 𝑖 − 1× 2−(1×1) − 𝑗 1×2−(2×1) + 𝑘 1×1−(2×−1) = 𝑖 −2−1 − 𝑗 2−2 + 𝑘 1+2 = − 3 𝒊 − 0 𝒋 + 3 𝒌 Magnitude of ( 𝑏1 × 𝑏2) = ( − 3)2+ 02+32 𝒃𝟏 × 𝒃𝟐 = 9+0+9 = 18 = 9 × 2 = 3 𝟐 Also, ( 𝑏1 × 𝑏2) . ( 𝑎2 – 𝑎1) = (− 3 𝑖−0 𝑗+3 𝑘).(1 𝑖 − 3 𝑗 − 2 𝑘) = (−3×1).(0×−3) + (3 × −2) = −3 − 0 − 6 = −9 So, shortest distance = 𝑏1 × 𝑏2 . 𝑎2 − 𝑎1 𝑏1 × 𝑏2 = − 93 2 = 3 2 = 3 2 × 2 2 = 𝟑 𝟐𝟐 Therefore, shortest distance between the given two lines is 3 22.

Chapter 11 Class 12 Three Dimensional Geometry

Serial order wise

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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.