Ex 11.2, 17 - Shortest distance r = (1-t)i + (t-2)j + (3-2t)k - Shortest distance between two skew lines

Slide52.JPG
Slide53.JPG

  1. Chapter 11 Class 12 Three Dimensional Geometry
  2. Serial order wise

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﷯ − 𝒂﷮𝟏﷯﷯) = (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﷯ 𝒃﷮𝟏﷯﷯× 𝒃﷮𝟐﷯﷯ ﷯ = 𝑖﷯﷮ 𝑗﷯﷮ 𝑘﷯﷮ −1﷮1﷮ −2﷮1﷮2﷮ −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﷯

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
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.