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An insect is crawling along the line insect is crawling along the line r = 6i Μ‚+2j Μ‚+2k Μ‚+Ξ»(i Μ‚-2j Μ‚+2k Μ‚ ) and another insect is crawling along the line r = βˆ’4i Μ‚-k Μ‚+ΞΌ(3i Μ‚-2j Μ‚-2k Μ‚ ). At what points on the lines should they reach so that the distance between them is the shortest? possible distance between them

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Question 34 (Choice 1) An insect is crawling along the line insect is crawling along the line π‘Ÿ βƒ— = 6𝑖 Μ‚+2𝑗 Μ‚+2π‘˜ Μ‚+πœ†(𝑖 Μ‚βˆ’2𝑗 Μ‚+2π‘˜ Μ‚ ) and another insect is crawling along the line π‘Ÿ βƒ— = βˆ’4𝑖 Μ‚βˆ’π‘˜ Μ‚+πœ‡(3𝑖 Μ‚βˆ’2𝑗 Μ‚βˆ’2π‘˜ Μ‚ ). 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 Line 2 π‘Ÿ βƒ— = βˆ’4𝑖 Μ‚βˆ’π‘˜ Μ‚ + πœ‡(3𝑖 Μ‚βˆ’2𝑗 Μ‚βˆ’2π‘˜ Μ‚ ) Line 1 π‘Ÿ βƒ— = 6𝑖 Μ‚+2𝑗 Μ‚+2π‘˜ Μ‚ +πœ†(𝑖 Μ‚βˆ’2𝑗 Μ‚+2π‘˜ Μ‚ ) 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 = 6𝑖 Μ‚+2𝑗 Μ‚+2π‘˜ Μ‚+πœ†(𝑖 Μ‚βˆ’2𝑗 Μ‚+2π‘˜ Μ‚ ) = (6+πœ†) 𝑖 Μ‚+(2βˆ’2πœ†)𝑗 Μ‚+(2+2πœ†)π‘˜ Μ‚ Point Q Since point Q lies on Line 2 Position vector of Q = βˆ’4𝑖 Μ‚βˆ’π‘˜ Μ‚+πœ‡(3𝑖 Μ‚βˆ’2𝑗 Μ‚βˆ’2π‘˜ Μ‚ ) = (βˆ’4+3πœ‡) 𝑖 Μ‚βˆ’2πœ‡π‘— Μ‚+(βˆ’1βˆ’2πœ‡)π‘˜ Μ‚ Now, (𝑷𝑸) βƒ— = Position vector of Q βˆ’ Position vector of P = [(βˆ’4+3πœ‡) 𝑖 Μ‚βˆ’2πœ‡π‘— Μ‚+(βˆ’1βˆ’2πœ‡) π‘˜ Μ‚ ]βˆ’[(6+πœ†) 𝑖 Μ‚+(2βˆ’2πœ†)𝑗 Μ‚+(2+2πœ†)π‘˜ Μ‚] = ο»Ώ(βˆ’10 + 3πœ‡ βˆ’ πœ†)πš€Μ‚ + (βˆ’2πœ‡ βˆ’ 2 + 2πœ†)πš₯Μ‚ + (βˆ’3 βˆ’ 2πœ‡ βˆ’ 2πœ†)π‘˜ Now, (𝑷𝑸) βƒ— βŠ₯ Line 1 (π‘Ÿ βƒ— = 6𝑖 Μ‚+2𝑗 Μ‚+2π‘˜ Μ‚+πœ†(𝑖 Μ‚βˆ’2𝑗 Μ‚+2π‘˜ Μ‚ )) Thus, (𝑃𝑄) βƒ— βŠ₯ (𝑖 Μ‚βˆ’2𝑗 Μ‚+2π‘˜ Μ‚ ) And (𝑷𝑸) βƒ— . (π’Š Μ‚βˆ’πŸπ’‹ Μ‚+πŸπ’Œ Μ‚ )=βˆ’πŸŽ (βˆ’10 + 3πœ‡ βˆ’ πœ†)πš€Μ‚ + (βˆ’2πœ‡ βˆ’ 2 + 2πœ†)πš₯Μ‚ + (βˆ’3 βˆ’ 2πœ‡ βˆ’ 2πœ†)π‘˜. (π’Š Μ‚βˆ’πŸπ’‹ Μ‚+πŸπ’Œ Μ‚ )=βˆ’πŸŽ (βˆ’10 + 3πœ‡ βˆ’ πœ†)1 + (βˆ’2πœ‡ βˆ’ 2 + 2πœ†)(βˆ’2) + (βˆ’3 βˆ’ 2πœ‡ βˆ’ 2πœ†)2 = 0 (βˆ’10 + 3πœ‡ βˆ’ πœ†) + (4πœ‡ + 4 βˆ’ 4πœ†) + (βˆ’6 βˆ’ 4πœ‡ βˆ’ 4πœ†) = 0 (βˆ’10 + 4 βˆ’ 6) + (βˆ’ πœ† βˆ’ 4πœ† βˆ’ 4πœ†) + (3πœ‡ + 4πœ‡ βˆ’ 4πœ‡) = 0 βˆ’12 βˆ’ 9πœ† + 3πœ‡ = 0 3πœ‡ βˆ’ 9πœ† = 12 3(πœ‡ βˆ’ 3πœ†) = 12 πœ‡ βˆ’ 3πœ† = 4 Similarly (𝑷𝑸) βƒ— βŠ₯ Line 2 (π‘Ÿ βƒ— = βˆ’4𝑖 Μ‚βˆ’π‘˜ Μ‚+πœ‡(3𝑖 Μ‚βˆ’2𝑗 Μ‚βˆ’2π‘˜ Μ‚ )) Thus, (𝑃𝑄) βƒ— βŠ₯(3𝑖 Μ‚βˆ’2𝑗 Μ‚βˆ’2π‘˜ Μ‚ ) And (𝑷𝑸) βƒ— .(3𝑖 Μ‚βˆ’2𝑗 Μ‚βˆ’2π‘˜ Μ‚ )=βˆ’πŸŽ (βˆ’10 + 3πœ‡ βˆ’ πœ†)πš€Μ‚ + (βˆ’2πœ‡ βˆ’ 2 + 2πœ†)πš₯Μ‚ + (βˆ’3 βˆ’ 2πœ‡ βˆ’ 2πœ†)π‘˜.(3𝑖 Μ‚βˆ’2𝑗 Μ‚βˆ’2π‘˜ Μ‚ )=βˆ’πŸŽ (βˆ’10 + 3πœ‡ βˆ’ πœ†)3 + (βˆ’2πœ‡ βˆ’ 2 + 2πœ†)(βˆ’2) + (βˆ’3 βˆ’ 2πœ‡ βˆ’ 2πœ†)(βˆ’2) = 0 (βˆ’30 + 9πœ‡ βˆ’ 3πœ†) + (4πœ‡ + 4 βˆ’ 4πœ†) + (6 + 4πœ‡ + 4πœ†) = 0 (βˆ’30 + 4 + 6) + (βˆ’3πœ† βˆ’ 4πœ† + 4πœ†) + (9πœ‡ + 4πœ‡ + 4πœ‡) = 0 βˆ’20 βˆ’ 3πœ† + 17πœ‡ = 0 17πœ‡ βˆ’ 3πœ† = 20 Thus, our equations are πœ‡ βˆ’ 3πœ† = 4 …(1) 17πœ‡ βˆ’ 3πœ† = 20 …(2) Solving (1) and (2) We get πœ‡ = 1, πœ† = βˆ’1 Point P Position vector of P = (6+πœ†) 𝑖 Μ‚+(2βˆ’2πœ†)𝑗 Μ‚+(2+2πœ†)π‘˜ Μ‚ Putting πœ† = βˆ’1 = (6+(βˆ’1)) 𝑖 Μ‚+(2βˆ’2(βˆ’1))𝑗 Μ‚+(2+2(βˆ’1))π‘˜ Μ‚ = πŸ“π’Š Μ‚+πŸ’π’‹ Μ‚ Point Q Position vector of Q = (βˆ’4+3πœ‡) 𝑖 Μ‚βˆ’2πœ‡π‘— Μ‚+(βˆ’1βˆ’2πœ‡)π‘˜ Μ‚ Putting πœ‡ = 1 = (βˆ’4+3(1)) 𝑖 Μ‚βˆ’2(1) 𝑗 Μ‚+(βˆ’1βˆ’2(1))π‘˜ Μ‚ = βˆ’π’Š Μ‚βˆ’πŸπ’‹ Μ‚βˆ’πŸ‘π‘˜ Μ‚Now, (𝑷𝑸) βƒ— = Position vector of Q βˆ’ Position vector of P = [βˆ’π‘– Μ‚βˆ’2𝑗 Μ‚βˆ’3π‘˜]βˆ’[5𝑖 Μ‚+4𝑗 Μ‚] = βˆ’πŸ”π’Š Μ‚βˆ’πŸ”π’‹ Μ‚βˆ’πŸ‘π’Œ Μ‚ And, Shortest distance = |(𝑃𝑄) βƒ— | = √((βˆ’6)^2+(βˆ’6)^2+(βˆ’3)^2 ) = √(36+36+9) = √81 = 9 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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.