Last updated at May 29, 2018 by Teachoo

Transcript

Misc 9 Find the shortest distance between lines 𝑟 = 6 𝑖 + 2 𝑗 + 2 𝑘 + 𝜆 ( 𝑖 – 2 𝑗 + 2 𝑘) and 𝑟 = –4 𝑖 – 𝑘 + 𝜇 (3 𝑖 – 2 𝑗 – 2 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) = (−4 𝑖 + 0 𝑗 − 1 𝑘) − (6 𝑖 + 2 𝑗 + 2 𝑘) = (−4 − 6) 𝑖 + (0 − 2) 𝑗 + (−1 − 2) 𝑘 = − 10 𝒊 − 2 𝒋 − 3 𝒌 ( 𝒃𝟏 × 𝒃𝟐) = 𝑖 𝑗 𝑘1 −223−2−2 = 𝑖 −2×−2−(−2×2) − 𝑗 1×−2−(3×2) + 𝑘 1×−2−(3×−2) = 𝑖 4+4 − 𝑗 −2−6 + 𝑘 −2+6 = 𝑖 (8) − 𝑗 (−8) + 𝑘(4) = 8 𝒊 + 8 𝒋 + 4 𝒌 Magnitude of 𝑏1 × 𝑏2 = 82+ 82+ 42 𝒃𝟏 × 𝒃𝟐 = 64+64+16 = 144 = 𝟏𝟐 Also, 𝒃𝟏× 𝒃𝟐 . 𝒂𝟐 − 𝒂𝟏 = (8 𝑖 + 8 𝑗 + 4 𝑘).(− 10 𝑖 − 2 𝑗 − 3 𝑘) = (8 × − 10) + (8 × − 2) + (4 × − 3) = − 80 + (−16) + (-12) = − 108 Shortest distance = 𝑏1 × 𝑏2 . 𝑎2 − 𝑎1 𝑏1 × 𝑏2 = −10812 = −9 = 9 Therefore, the shortest distance between the given two lines is 9.

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.