Example 20 - Equation of plane passing through intersection

Example 20 - Chapter 11 Class 12 Three Dimensional Geometry - Part 2
Example 20 - Chapter 11 Class 12 Three Dimensional Geometry - Part 3 Example 20 - Chapter 11 Class 12 Three Dimensional Geometry - Part 4

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Question 10 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 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.