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Find the shortest distance between the lines r β†’ Β  = 3iΒ  + 2jΒ  - 4kΒ  + Ξ» (iΒ  + 2jΒ  -2k )

And r = 5iΒ  + 2jΒ  - u (3iΒ  + 2jΒ  +6k). If the lines intersect find their point of intersection

Find shortest distance between lines r = 3i + 2j - 4k + (i + 2j + 2k)
Question 37 (Choice 1) - CBSE Class 12 Sample Paper for 2021 Boards - Part 2
Question 37 (Choice 1) - CBSE Class 12 Sample Paper for 2021 Boards - Part 3
Question 37 (Choice 1) - CBSE Class 12 Sample Paper for 2021 Boards - Part 4
Question 37 (Choice 1) - CBSE Class 12 Sample Paper for 2021 Boards - Part 5
Question 37 (Choice 1) - CBSE Class 12 Sample Paper for 2021 Boards - Part 6

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Note : This is similar to Ex 11.2, 14 ; Ex 11.2, 16 and Misc 9 of NCERT – Chapter 11 Class 12 Three Dimensional Geometry

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Transcript

Question 37 (Choice 1) Find the shortest distance between the lines π‘Ÿ βƒ— = 3𝑖 Μ‚ + 2𝑗 Μ‚ βˆ’ 4π‘˜ Μ‚ + πœ† (𝑖 Μ‚ + 2𝑗 Μ‚ + 2π‘˜ Μ‚) And π‘Ÿ βƒ— = 5𝑖 Μ‚ βˆ’ 2𝑗 Μ‚ + πœ‡(3𝑖 Μ‚ + 2𝑗 Μ‚ + 6π‘˜ Μ‚) If the lines intersect find their point of intersection 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— = (3π’Š Μ‚ + 2𝒋 Μ‚ βˆ’ 4π’Œ Μ‚) + πœ† (π’Š Μ‚ + 2𝒋 Μ‚ + 2π’Œ Μ‚) Comparing with π‘Ÿ βƒ— = (π‘Ž1) βƒ— + πœ†(𝑏1) βƒ— , (π‘Ž1) βƒ— = 3𝑖 Μ‚ + 2𝑗 Μ‚ βˆ’ 4π‘˜ Μ‚ & (𝑏1) βƒ— = 1𝑖 Μ‚ + 2𝑗 Μ‚ + 2π‘˜ Μ‚ 𝒓 βƒ— = (5π’Š Μ‚ βˆ’ 2𝒋 Μ‚) + 𝝁 (3π’Š Μ‚ + 2𝒋 Μ‚ + 6π’Œ Μ‚) Comparing with π‘Ÿ βƒ— = (π‘Ž2) βƒ— + πœ‡(𝑏2) βƒ— , (π‘Ž2) βƒ— = 5𝑖 Μ‚ βˆ’ 2𝑗 Μ‚ + 0π‘˜ Μ‚ & (𝑏2) βƒ— = 3𝑖 Μ‚ + 2𝑗 Μ‚ + 6π‘˜ Μ‚ Now, ((π’‚πŸ) βƒ— βˆ’ (π’‚πŸ) βƒ—) = (5𝑖 Μ‚ βˆ’ 2𝑗 Μ‚ + 0π‘˜ Μ‚) βˆ’ (3𝑖 Μ‚ + 2𝑗 Μ‚ βˆ’ 4π‘˜ Μ‚) = (5 βˆ’ 3) 𝑖 Μ‚ + (βˆ’2 βˆ’ 2)𝑗 Μ‚ + (0 βˆ’ (βˆ’4)) π‘˜ Μ‚ = 2π’Š Μ‚ βˆ’ 4𝒋 Μ‚ + 4π’Œ Μ‚ ((π’ƒπŸ) βƒ— Γ— (π’ƒπŸ) βƒ—) = |β– 8(𝑖 Μ‚&𝑗 Μ‚&π‘˜ Μ‚@1& 2&2@3&2&6)| = 𝑖 Μ‚ [(2 Γ— 6)βˆ’(2 Γ— 2)] βˆ’ 𝑗 Μ‚ [(1 Γ— 6)βˆ’(3 Γ— 2)] + π‘˜ Μ‚ [(1 Γ— 2)βˆ’(3 Γ— 2)] = 𝑖 Μ‚ [ 12βˆ’4] βˆ’ 𝑗 Μ‚ [6βˆ’6] + π‘˜ Μ‚ [2βˆ’6] = 𝑖 Μ‚ (8) βˆ’ 𝑗 Μ‚ (0) + π‘˜ Μ‚(βˆ’4) = 8π’Š Μ‚ βˆ’ 4π’Œ Μ‚ Magnitude of (𝑏1) βƒ— Γ— (𝑏2) βƒ— = √(8^2+0^2+γ€–(βˆ’4)γ€—^2 ) |(π’ƒπŸ) βƒ— Γ— (π’ƒπŸ) βƒ— | = √(64+16) = √80 = 4√5 Also, ((π’ƒπŸ) βƒ—Γ—(π’ƒπŸ) βƒ— ) . ((π’‚πŸ) βƒ— βˆ’ (π’‚πŸ) βƒ— ) = (8𝑖 Μ‚ βˆ’ 4π‘˜ Μ‚). (2𝑖 Μ‚ βˆ’ 4𝑗 Μ‚ + 4π‘˜ Μ‚) = (8 Γ— 2) + (0 Γ— βˆ’4) + (βˆ’4 Γ— 4) = βˆ’16 + 0 + (βˆ’16) = 0 Shortest distance = |(((𝑏1) βƒ— Γ— (𝑏2) βƒ— ) . ((π‘Ž2) βƒ— βˆ’ (π‘Ž1) βƒ— ))/|(𝑏1) βƒ— Γ— (𝑏2) βƒ— | | = |( 0)/(4√5)| = 0 Therefore, the shortest distance between the given two lines is 0 Finding Point of Intersection Since our lines are π‘Ÿ βƒ— = 3𝑖 Μ‚ + 2𝑗 Μ‚ βˆ’ 4π‘˜ Μ‚ + πœ† (𝑖 Μ‚ + 2𝑗 Μ‚ + 2π‘˜ Μ‚) And π‘Ÿ βƒ— = 5𝑖 Μ‚ βˆ’ 2𝑗 Μ‚ + πœ‡(3𝑖 Μ‚ + 2𝑗 Μ‚ + 6π‘˜ Μ‚) Thus, 3π’Š Μ‚ + 2𝒋 Μ‚ βˆ’ 4π’Œ Μ‚ + 𝝀 (π’Š Μ‚ + 2𝒋 Μ‚ + 2π’Œ Μ‚) = 5π’Š Μ‚ βˆ’ 2𝒋 Μ‚ + 𝝁(3π’Š Μ‚ + 2𝒋 Μ‚ + 6π’Œ Μ‚) (3 + πœ†)𝑖 Μ‚ + (2 + 2πœ†) 𝑗 Μ‚ + (βˆ’4 + 2πœ†) π‘˜ Μ‚ = (5 + 3πœ‡)𝑖 Μ‚ + (βˆ’2 + 2πœ‡)𝑗 Μ‚ + 6πœ‡π‘˜ Μ‚ Thus, 3 + πœ† = 5 + 3πœ‡ 2 + 2πœ† = βˆ’2 + 2πœ‡ βˆ’4 + 2πœ† = 6πœ‡ Solving (1) and (2) πœ† βˆ’ 3πœ‡ = 5 βˆ’ 3 𝝀 βˆ’ 3𝝁 = 2 And, 2 + 2πœ† = βˆ’2 + 2πœ‡ 2πœ† βˆ’ 2πœ‡ = βˆ’4 𝝀 βˆ’ 𝝁 = βˆ’2 Solving both equations 𝝀=βˆ’πŸ’, 𝝁=βˆ’πŸ Putting 𝝀=βˆ’πŸ’ in 𝒓 βƒ— π‘Ÿ βƒ— = 3𝑖 Μ‚ + 2𝑗 Μ‚ βˆ’ 4π‘˜ Μ‚ + πœ† (𝑖 Μ‚ + 2𝑗 Μ‚ + 2π‘˜ Μ‚) π‘Ÿ βƒ— = 3𝑖 Μ‚ + 2𝑗 Μ‚ βˆ’ 4π‘˜ Μ‚ βˆ’ 4(𝑖 Μ‚ + 2𝑗 Μ‚ + 2π‘˜ Μ‚) π‘Ÿ βƒ— = 3𝑖 Μ‚ + 2𝑗 Μ‚ βˆ’ 4π‘˜ Μ‚ βˆ’ 4𝑖 Μ‚ βˆ’ 8𝑗 Μ‚ βˆ’ 8π‘˜ Μ‚ π‘Ÿ βƒ— = βˆ’π‘– Μ‚ βˆ’ 6𝑗 Μ‚ βˆ’ 12π‘˜ Μ‚ So, Point of intersection is (βˆ’1, βˆ’6, βˆ’12)

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.