Find the shortest distance between the lines r   = 3i  + 2j  - 4k  + λ (i  + 2j  -2k )

And r = 5i  + 2j  - u (3i  + 2j  +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

 

 

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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  1. Class 12
  2. Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Transcript

Question 37 (Choice 1) Find the shortest distance between the lines ๐‘Ÿ โƒ— = 3๐‘– ฬ‚ + 2๐‘— ฬ‚ โˆ’ 4๐‘˜ ฬ‚ + ๐œ† (๐‘– ฬ‚ + 2๐‘— ฬ‚ + 2๐‘˜ ฬ‚) And ๐‘Ÿ โƒ— = 5๐‘– ฬ‚ โˆ’ 2๐‘— ฬ‚ + ๐œ‡(3๐‘– ฬ‚ + 2๐‘— ฬ‚ + 6๐‘˜ ฬ‚) If the lines intersect find their point of intersection Shortest distance between lines with vector equations ๐‘Ÿ โƒ— = (๐‘Ž1) โƒ— + ๐œ† (๐‘1) โƒ— and ๐‘Ÿ โƒ— = (๐‘Ž2) โƒ— + ๐œ‡(๐‘2) โƒ— is |(((๐’ƒ๐Ÿ) โƒ— ร— (๐’ƒ๐Ÿ) โƒ— ).((๐’‚๐Ÿ) โƒ— โˆ’ (๐’‚๐Ÿ) โƒ— ))/|(๐’ƒ๐Ÿ) โƒ— ร— (๐’ƒ๐Ÿ) โƒ— | | ๐’“ โƒ— = (3๐’Š ฬ‚ + 2๐’‹ ฬ‚ โˆ’ 4๐’Œ ฬ‚) + ๐œ† (๐’Š ฬ‚ + 2๐’‹ ฬ‚ + 2๐’Œ ฬ‚) Comparing with ๐‘Ÿ โƒ— = (๐‘Ž1) โƒ— + ๐œ†(๐‘1) โƒ— , (๐‘Ž1) โƒ— = 3๐‘– ฬ‚ + 2๐‘— ฬ‚ โˆ’ 4๐‘˜ ฬ‚ & (๐‘1) โƒ— = 1๐‘– ฬ‚ + 2๐‘— ฬ‚ + 2๐‘˜ ฬ‚ ๐’“ โƒ— = (5๐’Š ฬ‚ โˆ’ 2๐’‹ ฬ‚) + ๐ (3๐’Š ฬ‚ + 2๐’‹ ฬ‚ + 6๐’Œ ฬ‚) Comparing with ๐‘Ÿ โƒ— = (๐‘Ž2) โƒ— + ๐œ‡(๐‘2) โƒ— , (๐‘Ž2) โƒ— = 5๐‘– ฬ‚ โˆ’ 2๐‘— ฬ‚ + 0๐‘˜ ฬ‚ & (๐‘2) โƒ— = 3๐‘– ฬ‚ + 2๐‘— ฬ‚ + 6๐‘˜ ฬ‚ Now, ((๐’‚๐Ÿ) โƒ— โˆ’ (๐’‚๐Ÿ) โƒ—) = (5๐‘– ฬ‚ โˆ’ 2๐‘— ฬ‚ + 0๐‘˜ ฬ‚) โˆ’ (3๐‘– ฬ‚ + 2๐‘— ฬ‚ โˆ’ 4๐‘˜ ฬ‚) = (5 โˆ’ 3) ๐‘– ฬ‚ + (โˆ’2 โˆ’ 2)๐‘— ฬ‚ + (0 โˆ’ (โˆ’4)) ๐‘˜ ฬ‚ = 2๐’Š ฬ‚ โˆ’ 4๐’‹ ฬ‚ + 4๐’Œ ฬ‚ ((๐’ƒ๐Ÿ) โƒ— ร— (๐’ƒ๐Ÿ) โƒ—) = |โ– 8(๐‘– ฬ‚&๐‘— ฬ‚&๐‘˜ ฬ‚@1& 2&2@3&2&6)| = ๐‘– ฬ‚ [(2 ร— 6)โˆ’(2 ร— 2)] โˆ’ ๐‘— ฬ‚ [(1 ร— 6)โˆ’(3 ร— 2)] + ๐‘˜ ฬ‚ [(1 ร— 2)โˆ’(3 ร— 2)] = ๐‘– ฬ‚ [ 12โˆ’4] โˆ’ ๐‘— ฬ‚ [6โˆ’6] + ๐‘˜ ฬ‚ [2โˆ’6] = ๐‘– ฬ‚ (8) โˆ’ ๐‘— ฬ‚ (0) + ๐‘˜ ฬ‚(โˆ’4) = 8๐’Š ฬ‚ โˆ’ 4๐’Œ ฬ‚ Magnitude of (๐‘1) โƒ— ร— (๐‘2) โƒ— = โˆš(8^2+0^2+ใ€–(โˆ’4)ใ€—^2 ) |(๐’ƒ๐Ÿ) โƒ— ร— (๐’ƒ๐Ÿ) โƒ— | = โˆš(64+16) = โˆš80 = 4โˆš5 Also, ((๐’ƒ๐Ÿ) โƒ—ร—(๐’ƒ๐Ÿ) โƒ— ) . ((๐’‚๐Ÿ) โƒ— โˆ’ (๐’‚๐Ÿ) โƒ— ) = (8๐‘– ฬ‚ โˆ’ 4๐‘˜ ฬ‚). (2๐‘– ฬ‚ โˆ’ 4๐‘— ฬ‚ + 4๐‘˜ ฬ‚) = (8 ร— 2) + (0 ร— โˆ’4) + (โˆ’4 ร— 4) = โˆ’16 + 0 + (โˆ’16) = 0 Shortest distance = |(((๐‘1) โƒ— ร— (๐‘2) โƒ— ) . ((๐‘Ž2) โƒ— โˆ’ (๐‘Ž1) โƒ— ))/|(๐‘1) โƒ— ร— (๐‘2) โƒ— | | = |( 0)/(4โˆš5)| = 0 Therefore, the shortest distance between the given two lines is 0 Finding Point of Intersection Since our lines are ๐‘Ÿ โƒ— = 3๐‘– ฬ‚ + 2๐‘— ฬ‚ โˆ’ 4๐‘˜ ฬ‚ + ๐œ† (๐‘– ฬ‚ + 2๐‘— ฬ‚ + 2๐‘˜ ฬ‚) And ๐‘Ÿ โƒ— = 5๐‘– ฬ‚ โˆ’ 2๐‘— ฬ‚ + ๐œ‡(3๐‘– ฬ‚ + 2๐‘— ฬ‚ + 6๐‘˜ ฬ‚) Thus, 3๐’Š ฬ‚ + 2๐’‹ ฬ‚ โˆ’ 4๐’Œ ฬ‚ + ๐€ (๐’Š ฬ‚ + 2๐’‹ ฬ‚ + 2๐’Œ ฬ‚) = 5๐’Š ฬ‚ โˆ’ 2๐’‹ ฬ‚ + ๐(3๐’Š ฬ‚ + 2๐’‹ ฬ‚ + 6๐’Œ ฬ‚) (3 + ๐œ†)๐‘– ฬ‚ + (2 + 2๐œ†) ๐‘— ฬ‚ + (โˆ’4 + 2๐œ†) ๐‘˜ ฬ‚ = (5 + 3๐œ‡)๐‘– ฬ‚ + (โˆ’2 + 2๐œ‡)๐‘— ฬ‚ + 6๐œ‡๐‘˜ ฬ‚ Thus, 3 + ๐œ† = 5 + 3๐œ‡ 2 + 2๐œ† = โˆ’2 + 2๐œ‡ โˆ’4 + 2๐œ† = 6๐œ‡ Solving (1) and (2) ๐œ† โˆ’ 3๐œ‡ = 5 โˆ’ 3 ๐€ โˆ’ 3๐ = 2 And, 2 + 2๐œ† = โˆ’2 + 2๐œ‡ 2๐œ† โˆ’ 2๐œ‡ = โˆ’4 ๐€ โˆ’ ๐ = โˆ’2 Solving both equations ๐€=โˆ’๐Ÿ’, ๐=โˆ’๐Ÿ Putting ๐€=โˆ’๐Ÿ’ in ๐’“ โƒ— ๐‘Ÿ โƒ— = 3๐‘– ฬ‚ + 2๐‘— ฬ‚ โˆ’ 4๐‘˜ ฬ‚ + ๐œ† (๐‘– ฬ‚ + 2๐‘— ฬ‚ + 2๐‘˜ ฬ‚) ๐‘Ÿ โƒ— = 3๐‘– ฬ‚ + 2๐‘— ฬ‚ โˆ’ 4๐‘˜ ฬ‚ โˆ’ 4(๐‘– ฬ‚ + 2๐‘— ฬ‚ + 2๐‘˜ ฬ‚) ๐‘Ÿ โƒ— = 3๐‘– ฬ‚ + 2๐‘— ฬ‚ โˆ’ 4๐‘˜ ฬ‚ โˆ’ 4๐‘– ฬ‚ โˆ’ 8๐‘— ฬ‚ โˆ’ 8๐‘˜ ฬ‚ ๐‘Ÿ โƒ— = โˆ’๐‘– ฬ‚ โˆ’ 6๐‘— ฬ‚ โˆ’ 12๐‘˜ ฬ‚ So, Point of intersection is (โˆ’1, โˆ’6, โˆ’12)

Class 12
Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.