Last updated at May 29, 2018 by Teachoo

Transcript

Ex 11.2, 17 Find the shortest distance between the lines whose vector equations are 𝑟 = (1 − t) 𝑖 + (t − 2) 𝑗 + (3 − 2t) 𝑘 and 𝑟 = (s + 1) 𝑖 + (2s – 1) 𝑗 – (2s + 1) 𝑘 Shortest distance between lines with vector equations 𝑟 = 𝑎1 + 𝜆 𝑏1 and 𝑟 = 𝑎2 + 𝜇 𝑏2 is ( 𝑏1× 𝑏2).( 𝑎2 − 𝑎1) 𝑏1 × 𝑏2 Now, ( 𝒂𝟐 − 𝒂𝟏) = (1 𝑖 − 1 𝑗 − 1 𝑘) − (1 𝑖 − 2𝑗 + 3 𝑘) = (1 − 1) 𝑖 + ( − 1 + 2) 𝑗 + ( − 1 − 3) 𝑘 = 0 𝒊 + 1 𝒋 − 4 𝒌 𝒃𝟏× 𝒃𝟐 = 𝑖 𝑗 𝑘 −11 −212 −2 = 𝑖 1×− 2 − 2×− 2 − 𝑗 −1×−2 1×− 2 + 𝑘 − 1×2− 1×1 = 𝑖 − 2+4 − 𝑗 2+2 A + 𝑘 −2−1 = 2 𝒊 − 4 𝒋 − 3 𝒌 Magnitude of ( 𝑏1× 𝑏2) = 22+(− 4)2+(− 3)2 𝒃𝟏× 𝒃𝟐 = 4+16+9 = 𝟐𝟗

Ex 11.2

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Ex 11.2, 17 Important You are here

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 9 years. He provides courses for Maths and Science at Teachoo.