Check sibling questions

Misc 9 - Find shortest distance between lines r = 6i + 2j + 2k + (

Misc 9 - Chapter 11 Class 12 Three Dimensional Geometry - Part 2
Misc 9 - Chapter 11 Class 12 Three Dimensional Geometry - Part 3 Misc 9 - Chapter 11 Class 12 Three Dimensional Geometry - Part 4

This video is only available for Teachoo black users

Get live Maths 1-on-1 Classs - Class 6 to 12


Transcript

Misc 9 Find the shortest distance between lines π‘Ÿ βƒ— = 6𝑖 Μ‚ + 2𝑗 Μ‚ + 2π‘˜ Μ‚ + πœ† (𝑖 Μ‚ – 2𝑗 Μ‚ + 2π‘˜ Μ‚) and π‘Ÿ βƒ— = –4𝑖 Μ‚ – π‘˜ Μ‚ + πœ‡ (3𝑖 Μ‚ – 2𝑗 Μ‚ – 2π‘˜ Μ‚) .Shortest distance between lines with vector equations π‘Ÿ βƒ— = (π‘Ž1) βƒ— + πœ† (𝑏1) βƒ— and π‘Ÿ βƒ— = (π‘Ž2) βƒ— + πœ‡(𝑏2) βƒ— is |(((π’ƒπŸ) βƒ— Γ— (π’ƒπŸ) βƒ— ).((π’‚πŸ) βƒ— βˆ’ (π’‚πŸ) βƒ— ))/|(π’ƒπŸ) βƒ— Γ— (π’ƒπŸ) βƒ— | | 𝒓 βƒ— = (6π’Š Μ‚ + 2𝒋 Μ‚ + 2π’Œ Μ‚) + πœ† (π’Š Μ‚ βˆ’ 2𝒋 Μ‚ + 2π’Œ Μ‚) Comparing with π‘Ÿ βƒ— = (π‘Ž1) βƒ— + πœ†(𝑏1) βƒ— , (π‘Ž1) βƒ— = 6𝑖 Μ‚ + 2𝑗 Μ‚ + 2π‘˜ Μ‚ & (𝑏1) βƒ— = 1𝑖 Μ‚ βˆ’ 2𝑗 Μ‚ + 2π‘˜ Μ‚ 𝒓 βƒ— = (βˆ’4π’Š Μ‚ βˆ’ π’Œ Μ‚) + 𝝁 (3π’Š Μ‚ βˆ’ 2𝒋 Μ‚ βˆ’ 2π’Œ Μ‚) Comparing with π‘Ÿ βƒ— = (π‘Ž2) βƒ— + πœ‡(𝑏2) βƒ— , (π‘Ž2) βƒ— = βˆ’ 4𝑖 Μ‚ + 0𝑗 Μ‚ βˆ’ 1π‘˜ Μ‚ & (𝑏2) βƒ— = 3𝑖 Μ‚ βˆ’ 2𝑗 Μ‚ βˆ’ 2π‘˜ Μ‚ Now, ((π’‚πŸ) βƒ— βˆ’ (π’‚πŸ) βƒ—) = (βˆ’4𝑖 Μ‚ + 0𝑗 Μ‚ βˆ’ 1π‘˜ Μ‚) βˆ’ (6𝑖 Μ‚ + 2𝑗 Μ‚ + 2π‘˜ Μ‚) = (βˆ’4 βˆ’ 6) 𝑖 Μ‚ + (0 βˆ’ 2)𝑗 Μ‚ + (βˆ’1 βˆ’ 2) π‘˜ Μ‚ = βˆ’ 10π’Š Μ‚ βˆ’ 2𝒋 Μ‚ βˆ’ 3π’Œ Μ‚ ((π’ƒπŸ) βƒ— Γ— (π’ƒπŸ) βƒ—) = |β– 8(𝑖 Μ‚&𝑗 Μ‚&π‘˜ Μ‚@1& βˆ’2&[email protected]&βˆ’2&βˆ’2)| = 𝑖 Μ‚ [(βˆ’2Γ—βˆ’2)βˆ’(βˆ’2Γ—2)] βˆ’ 𝑗 Μ‚ [(1Γ—βˆ’2)βˆ’(3Γ—2)] + π‘˜ Μ‚ [(1Γ—βˆ’2)βˆ’(3Γ—βˆ’2)] = 𝑖 Μ‚ [ 4+4] βˆ’ 𝑗 Μ‚ [βˆ’2βˆ’6] + π‘˜ Μ‚ [βˆ’2+6] = 𝑖 Μ‚ (8) βˆ’ 𝑗 Μ‚ (βˆ’8) + π‘˜ Μ‚(4) = 8π’Š Μ‚ + 8𝒋 Μ‚ + 4π’Œ Μ‚ Magnitude of (𝑏1) βƒ— Γ— (𝑏2) βƒ— = √(8^2+8^2+4^2 ) |(π’ƒπŸ) βƒ— Γ— (π’ƒπŸ) βƒ— | = √(64+64+16) = √144 = 𝟏𝟐 Also, ((π’ƒπŸ) βƒ—Γ—(π’ƒπŸ) βƒ— ) . ((π’‚πŸ) βƒ— βˆ’ (π’‚πŸ) βƒ— ) = (8𝑖 Μ‚ + 8𝑗 Μ‚ + 4π‘˜ Μ‚).(βˆ’ 10𝑖 Μ‚ βˆ’ 2𝑗 Μ‚ βˆ’ 3π‘˜ Μ‚) = (8 Γ— βˆ’ 10) + (8 Γ— βˆ’ 2) + (4 Γ— βˆ’ 3) = βˆ’ 80 + (βˆ’16) + (-12) = βˆ’ 108 Shortest distance = |(((𝑏1) βƒ— Γ— (𝑏2) βƒ— ) . ((π‘Ž2) βƒ— βˆ’ (π‘Ž1) βƒ— ))/|(𝑏1) βƒ— Γ— (𝑏2) βƒ— | | = |( βˆ’108)/12| = |βˆ’9| = 9 Therefore, the shortest distance between the given two lines is 9.

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.