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

Transcript

Ex 11.3, 10 Find the vector equation of plane passing through the intersection of the planes ๐‘Ÿ โƒ— . (2๐‘– ฬ‚ + 2๐‘— ฬ‚ โˆ’ 3๐‘˜ ฬ‚) = 7, ๐‘Ÿ โƒ— .(2๐‘– ฬ‚ + 5๐‘— ฬ‚ + 3๐‘˜ ฬ‚) = 9 and through the point (2, 1, 3).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 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—. (2๐’Š ฬ‚ + 2๐’‹ ฬ‚ โˆ’ 3๐’Œ ฬ‚) = 7 Comparing with ๐‘Ÿ โƒ—.(๐‘›1) โƒ— = ๐‘‘1, (๐‘›1) โƒ— = 2๐‘– ฬ‚ + 2๐‘— ฬ‚ โˆ’ 3๐‘˜ ฬ‚ & d1 = 7 ๐’“ โƒ—. (2๐’Š ฬ‚ + 5๐’‹ ฬ‚ + 3๐’Œ ฬ‚) = 9 Comparing with ๐‘Ÿ โƒ—.(๐‘›2) โƒ— = ๐‘‘2, (๐‘›2) โƒ— = 2๐‘– ฬ‚ + 5๐‘— ฬ‚ + 3๐‘˜ ฬ‚ & d2 = 9 So, equation of the plane is ๐‘Ÿ โƒ—.["(2" ๐‘– ฬ‚+"2" ๐‘— ฬ‚" " โˆ’"3" ๐‘˜ ฬ‚")" +"๐œ†(2" ๐‘– ฬ‚" " + 5๐‘— ฬ‚" " + "3" ๐‘˜ ฬ‚")" ] = 7 + ๐œ†.9 ๐‘Ÿ โƒ—. ["2" ๐‘– ฬ‚" " +" 2" ๐‘— ฬ‚" " โˆ’ "3" ๐‘˜ ฬ‚ + 2"๐œ†" ๐‘– ฬ‚ + 5"๐œ†" ๐‘— ฬ‚ + 3"๐œ†" ๐‘˜ ฬ‚ ] = 7 + 9"๐œ†" ๐’“ โƒ—. ["(2" +"2๐œ†" )๐’Š ฬ‚" " +"(2" +"5๐œ†" )๐’‹ ฬ‚ +"(โˆ’" ๐Ÿ‘+"3๐œ†" )๐’Œ ฬ‚ ] = 9"๐œ†" + 7 Now, to find ๐œ† , put ๐’“ โƒ— = x๐’Š ฬ‚ + y๐’‹ ฬ‚ + z๐’Œ ฬ‚ (x๐‘– ฬ‚ + y๐‘— ฬ‚ + z๐‘˜ ฬ‚).["(2 " + "2๐œ†" )๐‘– ฬ‚" " + "(2" +"5๐œ†" )๐‘— ฬ‚ +"(โˆ’" 3" " +" 3๐œ†" )๐‘˜ ฬ‚ ] = 9๐œ† + 7 x"(2 "+" 2๐œ†")" "+ "y (2 "+" 5๐œ†")๐‘— ฬ‚ + ๐‘ง"("โˆ’3+"3๐œ†")๐‘˜ ฬ‚ = 9๐œ† + 7 The plane passes through (2, 1, 3) Putting (2, 1, 3) in (2), 2(2 + 2๐œ†) + 1(2 + 5๐œ†) + 3(โˆ’3 + 3๐œ†) = 9๐œ† + 7 4 + 4๐œ† + 2 + 5๐œ† + (โˆ’9) + 9๐œ† = 9๐œ† + 7 18๐œ† โˆ’ 9๐œ† = 7 + 3 9๐œ† = 10 โˆด ๐œ† = ๐Ÿ๐ŸŽ/๐Ÿ— Putting value of ๐œ† in (1), ๐‘Ÿ โƒ—. [(2+ 2. 10/9) ๐‘– ฬ‚ + (2+ 5. 10/9) ๐‘— ฬ‚ +("โˆ’" 3+ 3. 10/9) ๐‘˜ ฬ‚ ] = 9.10/9 + 7 ๐‘Ÿ โƒ—. [(2 + 20/9) ๐‘– ฬ‚ +(2 + 50/9) ๐‘— ฬ‚ +("โˆ’" 3 + 30/9) ๐‘˜ ฬ‚ ] = 10 + 7 ๐‘Ÿ โƒ—. [38/9 ๐‘– ฬ‚+ 68/9 ๐‘— ฬ‚ + 3/9 ๐‘˜ ฬ‚ ] = 17 1/9 ๐‘Ÿ โƒ—. (38๐‘– ฬ‚ + 68๐‘— ฬ‚ + 3๐‘˜ ฬ‚ ) = 17 ๐‘Ÿ โƒ—.(38๐‘– ฬ‚+68๐‘— ฬ‚+3๐‘˜ ฬ‚ ) = 17 ร— 9 ๐’“ โƒ—.(๐Ÿ‘๐Ÿ–๐’Š ฬ‚+๐Ÿ”๐Ÿ–๐’‹ ฬ‚+๐Ÿ‘๐’Œ ฬ‚ ) = 153

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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.