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  1. Chapter 11 Class 12 Three Dimensional Geometry
  2. Serial order wise

Transcript

Example 20 Find the vector equation of the plane passing through the intersection of the planes ๐‘Ÿ โƒ— . (๐‘– ฬ‚ + ๐‘— ฬ‚ + ๐‘˜ ฬ‚) = 6 and ๐‘Ÿ โƒ— . (2๐‘– ฬ‚ + 3๐‘— ฬ‚ + 4๐‘˜ ฬ‚) = โˆ’ 5, and the point (1, 1, 1).The vector equation of a plane passing through the intersection of planes ๐‘Ÿ โƒ—. (๐‘›1) โƒ— = d1 and ๐‘Ÿ โƒ—. (๐‘›2) โƒ— = d2 and also through the point (x1, y1, z1) is ๐’“ โƒ—.((๐’๐Ÿ) โƒ— + ๐œ†(๐’๐Ÿ) โƒ—) = d1 + ๐œ†d2 Given, the plane passes through ๐’“ โƒ—.(๐’Š ฬ‚ + ๐’‹ ฬ‚ + ๐’Œ ฬ‚) = 6 Comparing with ๐‘Ÿ โƒ—.(๐‘›1) โƒ— = d1, (๐’๐Ÿ) โƒ— = ๐’Š ฬ‚ + ๐’‹ ฬ‚ + ๐’Œ ฬ‚ & d1 = 6 ๐’“ โƒ—.(2๐’Š ฬ‚ + 3๐’‹ ฬ‚ + 4๐’Œ ฬ‚) = โˆ’5 โ€“๐‘Ÿ โƒ—.(2๐‘– ฬ‚ + 3๐‘— ฬ‚ + 4๐‘˜ ฬ‚) = 5 ๐‘Ÿ โƒ— .(โˆ’ 2๐‘– ฬ‚ โˆ’ 3๐‘— ฬ‚ โˆ’ 4๐‘˜ ฬ‚) = 5 Comparing with ๐‘Ÿ โƒ—.(๐‘›2) โƒ— = d2 (๐’๐Ÿ) โƒ— = โˆ’ 2๐’Š ฬ‚ โˆ’ 3๐’‹ ฬ‚ โˆ’ 4๐’Œ ฬ‚ & d2 = 5 Equation of plane is ๐‘Ÿ โƒ—. [(๐‘– ฬ‚+๐‘— ฬ‚+๐‘˜ ฬ‚ )+"๐œ†" (โˆ’2๐‘– ฬ‚โˆ’3๐‘— ฬ‚โˆ’4๐‘˜ ฬ‚)] = 6 + ๐œ†5 ๐’“ โƒ—. [(๐’Š ฬ‚" " +๐’‹ ฬ‚" " +๐’Œ ฬ‚ )โˆ’"๐œ†" (๐Ÿ๐’Š ฬ‚+๐Ÿ‘๐’‹ ฬ‚+๐Ÿ’๐’Œ ฬ‚)] = 6 + 5๐œ† Now to find ๐œ† , put ๐’“ โƒ— = x๐’Š ฬ‚ + y๐’‹ ฬ‚ + z๐’Œ ฬ‚ (x๐‘– ฬ‚ + y๐‘— ฬ‚ + z๐‘˜ ฬ‚). [(๐‘– ฬ‚+๐‘— ฬ‚+๐‘˜ ฬ‚ )โˆ’"๐œ†" (2๐‘– ฬ‚+3๐‘— ฬ‚+4๐‘˜ ฬ‚)] = 5๐œ† + 6 (x๐‘– ฬ‚ + y๐‘— ฬ‚ + z๐‘˜ ฬ‚).(๐‘– ฬ‚+๐‘— ฬ‚+๐‘˜ ฬ‚ ) โˆ’ ๐œ† (x๐‘– ฬ‚ + y๐‘— ฬ‚ + z๐‘˜ ฬ‚).(2๐‘– ฬ‚+3๐‘— ฬ‚+4๐‘˜ ฬ‚) = 5๐œ† + 6 (x ร— 1) + (y ร— 1) + (z ร— 1) โˆ’ ๐œ†[(๐‘ฅร—2)+(๐‘ฆร—3)+(๐‘งร—4)] = 5๐œ† + 6 x + y + z โˆ’ ๐œ†[2๐‘ฅ+3๐‘ฆ+4๐‘ง] = 5๐œ† + 6 x + y + z โˆ’ 2๐œ†๐‘ฅ โˆ’ 3๐œ†y โˆ’ 4๐œ†z = 5๐œ† + 6 (1 โˆ’ 2๐œ†)x + (1 โˆ’ 3๐œ†)y + (1 โˆ’ 4๐œ†) z = 5๐œ† + 6 Since the plane passes through (1, 1, 1), Putting (1, 1, 1) in (2) (1 โˆ’ 2๐œ†)x + (1 โˆ’ 3๐œ†)y + (1 โˆ’ 4๐œ†) z = 5๐œ† + 6 (1 โˆ’2๐œ†) ร— 1 + (1 โˆ’ 3๐œ†) ร— 1 + (1 โˆ’ 4๐œ†) ร— 1 = 5๐œ† + 6 1 โˆ’2๐œ† + 1 โˆ’ 3๐œ† + 1 โˆ’ 4๐œ†= 5๐œ† + 6 3 โˆ’ 9๐œ† = 5๐œ† + 6 โˆ’14๐œ† = 3 โˆด ๐œ† = (โˆ’๐Ÿ‘)/๐Ÿ๐Ÿ’ Putting value of ๐œ† in (1), ๐‘Ÿ โƒ—. [(๐‘– ฬ‚" " +" " ๐‘— ฬ‚" " +" " ๐‘˜ ฬ‚ )โˆ’(( โˆ’3)/14)(2๐‘– ฬ‚+3๐‘— ฬ‚+"4" ๐‘˜ ฬ‚)]= 6 + 5 ร— ( โˆ’3)/14 ๐‘Ÿ โƒ—. [(๐‘– ฬ‚+๐‘— ฬ‚+" " ๐‘˜ ฬ‚ )+3/14(2๐‘– ฬ‚+3๐‘— ฬ‚+"4" ๐‘˜ ฬ‚)]= 6 โˆ’ 15/14 ๐‘Ÿ โƒ—. [๐‘– ฬ‚+๐‘— ฬ‚" " +๐‘˜ ฬ‚+ 6/14 ๐‘– ฬ‚+9/14 ๐‘— ฬ‚+12/14 ๐‘˜ ฬ‚ ]= 69/14 ๐‘Ÿ โƒ—. [(1+6/14) ๐‘– ฬ‚ +(1+9/14) ๐‘— ฬ‚+(1+12/14) ๐‘˜ ฬ‚ ]= 69/14 ๐‘Ÿ โƒ—. [20/14 ๐‘– ฬ‚ + 23/14 ๐‘— ฬ‚ + 26/14 ๐‘˜ ฬ‚ ]= 69/14 ๐‘Ÿ โƒ—. [1/14(20๐‘– ฬ‚+23๐‘— ฬ‚+26๐‘˜ ฬ‚)]= 69/14 1/14 ๐‘Ÿ โƒ—. (20๐‘– ฬ‚ + 23๐‘— ฬ‚ + 26๐‘˜ ฬ‚) = 69/14 ๐‘Ÿ โƒ—. (20๐‘– ฬ‚ + 23๐‘— ฬ‚ + 26๐‘˜ ฬ‚) = 69 Therefore, the vector equation of the required plane is ๐’“ โƒ—.(๐Ÿ๐ŸŽ๐’Š ฬ‚ + ๐Ÿ๐Ÿ‘๐’‹ ฬ‚ + ๐Ÿ๐Ÿ”๐’Œ ฬ‚) = ๐Ÿ”๐Ÿ—

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.