An aeroplane is flying along the line r =λ( i  ˆ - j  ˆ + k  ˆ ); where ' λ ' is a scalar and another aeroplane is flying along the line r =i ˆ-j ˆ+μ(-2j ˆ+ k  ˆ ); where ' μ ' is a scalar. At what points on the lines should they reach, so that the distance between them is the shortest? Find the shortest possible distance between them.

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ο»ΏThe given lines are non-parallel lines. Let Shortest distance = |(𝑃𝑄) βƒ— | Since (𝑃𝑄) βƒ— is shortest distance, (𝑃𝑄) βƒ— βŠ₯ Line 1 (𝑃𝑄) βƒ— βŠ₯ Line 2 Point P Since point P lies on Line 1 Position vector of P = πœ†(𝑖 Λ†βˆ’π‘— Λ†+π‘˜ Λ†) = (πœ†) 𝑖 Μ‚+(βˆ’πœ†)𝑗 Μ‚+(πœ†)π‘˜ Μ‚ Now, (𝑷𝑸) βƒ— = Position vector of Q βˆ’ Position vector of P = [𝑖 Μ‚+(βˆ’1βˆ’2πœ‡)𝑗 Λ†+(πœ‡)π‘˜ Μ‚ ]βˆ’[(πœ†) 𝑖 Μ‚+(βˆ’πœ†)𝑗 Μ‚+(πœ†)π‘˜ Μ‚] = ο»Ώ(1 βˆ’ πœ†)πš€Μ‚ + (βˆ’ 1 βˆ’ 2πœ‡ + πœ†)πš₯Μ‚ + (πœ‡ βˆ’ πœ†)π‘˜ Μ‚ Now, (𝑷𝑸) βƒ— βŠ₯ Line 1 (π‘Ÿ βƒ— = πœ†(𝑖 Λ†βˆ’π‘— Λ†+π‘˜ Λ†) ) Thus, (𝑃𝑄) βƒ— βŠ₯ (𝑖 Μ‚βˆ’π‘— Μ‚+π‘˜ Μ‚ ) And (𝑷𝑸) βƒ— . (π’Š Μ‚βˆ’π’‹ Μ‚+π’Œ Μ‚ )=𝟎 ο»Ώ(1 βˆ’ πœ†)πš€Μ‚ + (βˆ’ 1 βˆ’ 2πœ‡ + πœ†)πš₯Μ‚ + (πœ‡ βˆ’ πœ†)π‘˜ Μ‚. (π’Š Μ‚βˆ’π’‹ Μ‚+π’Œ Μ‚ )=𝟎 (1βˆ’ πœ†)1 + (βˆ’1 βˆ’2πœ‡ + πœ†)(βˆ’1) + (πœ‡ βˆ’ πœ†)1 = 0 (1βˆ’ πœ†) + (1 + 2πœ‡ - πœ†) + (πœ‡ βˆ’ πœ†) = 0 (1 + 1) + (βˆ’ πœ† βˆ’ πœ† βˆ’ πœ†) + (2πœ‡ + πœ‡) = 0 2 βˆ’ 3πœ† + 3πœ‡ = 0 3πœ‡ βˆ’ 3πœ† = βˆ’ 2 Similarly (𝑷𝑸) βƒ— βŠ₯ Line 2 (π‘Ÿ βƒ— = 𝑖 Λ†βˆ’π‘— Λ†+πœ‡(βˆ’2𝑗 Λ†+π‘˜ Λ†)) Thus, (𝑃𝑄) βƒ— βŠ₯(βˆ’2𝑗 Λ†+π‘˜ Λ†) And (𝑷𝑸) βƒ— .(βˆ’2𝑗 Λ†+π‘˜ Λ†)=𝟎 (1 βˆ’ πœ†)πš€Μ‚ + (βˆ’ 1 βˆ’ 2πœ‡ + πœ†)πš₯Μ‚ + (πœ‡ βˆ’ πœ†)π‘˜ Μ‚.(βˆ’2𝑗 Λ†+π‘˜ Λ† )=0 (1 βˆ’ πœ†)(0) + (βˆ’ 1 βˆ’ 2πœ‡ + πœ†)(βˆ’2) + (πœ‡ βˆ’ πœ†)(1) = 0 0 + (2 +4πœ‡ βˆ’ 2πœ†) + (πœ‡ βˆ’ πœ†) = 0 (2) + (βˆ’2πœ† βˆ’ πœ†) + (4πœ‡ + πœ‡) = 0 2 βˆ’ 3πœ† + 5πœ‡ = 0 5πœ‡ βˆ’ 3πœ† = βˆ’2 Thus, our equations are 3πœ‡ βˆ’ 3πœ† = βˆ’2 …(1) 5πœ‡ βˆ’ 3πœ† = βˆ’2 …(2) Solving (1) and (2) We get πœ‡ = 0, πœ† = 𝟐/πŸ‘ Point P Position vector of P = (πœ†) 𝑖 Μ‚+(βˆ’πœ†)𝑗 Μ‚+(πœ†)π‘˜ Μ‚ Putting πœ† = 𝟐/πŸ‘ = 𝟐/πŸ‘ π’Š Μ‚βˆ’πŸ/πŸ‘ 𝒋 Μ‚+ 𝟐/πŸ‘ π’Œ Μ‚ Point Q Position vector of Q = 𝑖 Μ‚+(βˆ’1βˆ’2πœ‡)𝑗 Λ†+(πœ‡)π‘˜ Μ‚ Putting πœ‡ = 0 =( 𝑖) Μ‚+(βˆ’1 βˆ’0) 𝑗 Μ‚+(0)π‘˜ Μ‚ = π’Š Μ‚βˆ’π’‹ Μ‚ Now, (𝑷𝑸) βƒ— = Position vector of Q βˆ’ Position vector of P =[ π’Š Μ‚βˆ’(𝒋]) Μ‚βˆ’[ 𝟐/πŸ‘ π’Š Μ‚βˆ’πŸ/πŸ‘ 𝒋 Μ‚+ 𝟐/πŸ‘ π’Œ Μ‚] = 𝟏/πŸ‘ π’Š Μ‚βˆ’(𝟏 )/πŸ‘ 𝒋 Μ‚βˆ’πŸ/πŸ‘ π’Œ Μ‚ And, Shortest distance = |(𝑃𝑄) βƒ— | = √((1/3)^2+(βˆ’1/3)^2+(2/3)^2 ) =√( 1/9+1/9 +4/9) = √(6/9) = √(𝟐/πŸ‘) units

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo