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Chapter 11 Class 12 Three Dimensional Geometry
Serial order wise

Ex 11.3, 10 - Find vector equation of plane passing through intersect

Ex 11.3, 10 - Chapter 11 Class 12 Three Dimensional Geometry - Part 2
Ex 11.3, 10 - Chapter 11 Class 12 Three Dimensional Geometry - Part 3

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Transcript

Ex 11.3, 10 Find the vector equation of plane passing through the intersection of the planes π‘Ÿ βƒ— . (2𝑖 Μ‚ + 2𝑗 Μ‚ βˆ’ 3π‘˜ Μ‚) = 7, π‘Ÿ βƒ— .(2𝑖 Μ‚ + 5𝑗 Μ‚ + 3π‘˜ Μ‚) = 9 and through the point (2, 1, 3).The vector equation of a plane passing through the intersection of planes π‘Ÿ βƒ—. (𝑛1) βƒ— = d1 and π‘Ÿ βƒ—. (𝑛2) βƒ— = d2 and also through the point (x1, y1, z1) is 𝒓 βƒ—.((π’πŸ) βƒ— + πœ†(π’πŸ) βƒ—) = d1 + πœ†d2 Given, the plane passes through The vector equation of a plane passing through the intersection of planes π‘Ÿ βƒ—. (𝑛1) βƒ— = d1 and π‘Ÿ βƒ—. (𝑛2) βƒ— = d2 and also through the point (x1, y1, z1) is 𝒓 βƒ—.((π’πŸ) βƒ— + πœ†(π’πŸ) βƒ—) = d1 + πœ†d2 Given, the plane passes through 𝒓 βƒ—. (2π’Š Μ‚ + 2𝒋 Μ‚ βˆ’ 3π’Œ Μ‚) = 7 Comparing with π‘Ÿ βƒ—.(𝑛1) βƒ— = 𝑑1, (𝑛1) βƒ— = 2𝑖 Μ‚ + 2𝑗 Μ‚ βˆ’ 3π‘˜ Μ‚ & d1 = 7 𝒓 βƒ—. (2π’Š Μ‚ + 5𝒋 Μ‚ + 3π’Œ Μ‚) = 9 Comparing with π‘Ÿ βƒ—.(𝑛2) βƒ— = 𝑑2, (𝑛2) βƒ— = 2𝑖 Μ‚ + 5𝑗 Μ‚ + 3π‘˜ Μ‚ & d2 = 9 So, equation of the plane is π‘Ÿ βƒ—.["(2" 𝑖 Μ‚+"2" 𝑗 Μ‚" " βˆ’"3" π‘˜ Μ‚")" +"πœ†(2" 𝑖 Μ‚" " + 5𝑗 Μ‚" " + "3" π‘˜ Μ‚")" ] = 7 + πœ†.9 π‘Ÿ βƒ—. ["2" 𝑖 Μ‚" " +" 2" 𝑗 Μ‚" " βˆ’ "3" π‘˜ Μ‚ + 2"πœ†" 𝑖 Μ‚ + 5"πœ†" 𝑗 Μ‚ + 3"πœ†" π‘˜ Μ‚ ] = 7 + 9"πœ†" 𝒓 βƒ—. ["(2" +"2πœ†" )π’Š Μ‚" " +"(2" +"5πœ†" )𝒋 Μ‚ +"(βˆ’" πŸ‘+"3πœ†" )π’Œ Μ‚ ] = 9"πœ†" + 7 Now, to find πœ† , put 𝒓 βƒ— = xπ’Š Μ‚ + y𝒋 Μ‚ + zπ’Œ Μ‚ (x𝑖 Μ‚ + y𝑗 Μ‚ + zπ‘˜ Μ‚).["(2 " + "2πœ†" )𝑖 Μ‚" " + "(2" +"5πœ†" )𝑗 Μ‚ +"(βˆ’" 3" " +" 3πœ†" )π‘˜ Μ‚ ] = 9πœ† + 7 x"(2 "+" 2πœ†")" "+ "y (2 "+" 5πœ†")𝑗 Μ‚ + 𝑧"("βˆ’3+"3πœ†")π‘˜ Μ‚ = 9πœ† + 7 The plane passes through (2, 1, 3) Putting (2, 1, 3) in (2), 2(2 + 2πœ†) + 1(2 + 5πœ†) + 3(βˆ’3 + 3πœ†) = 9πœ† + 7 4 + 4πœ† + 2 + 5πœ† + (βˆ’9) + 9πœ† = 9πœ† + 7 18πœ† βˆ’ 9πœ† = 7 + 3 9πœ† = 10 ∴ πœ† = 𝟏𝟎/πŸ— Putting value of πœ† in (1), π‘Ÿ βƒ—. [(2+ 2. 10/9) 𝑖 Μ‚ + (2+ 5. 10/9) 𝑗 Μ‚ +("βˆ’" 3+ 3. 10/9) π‘˜ Μ‚ ] = 9.10/9 + 7 π‘Ÿ βƒ—. [(2 + 20/9) 𝑖 Μ‚ +(2 + 50/9) 𝑗 Μ‚ +("βˆ’" 3 + 30/9) π‘˜ Μ‚ ] = 10 + 7 π‘Ÿ βƒ—. [38/9 𝑖 Μ‚+ 68/9 𝑗 Μ‚ + 3/9 π‘˜ Μ‚ ] = 17 1/9 π‘Ÿ βƒ—. (38𝑖 Μ‚ + 68𝑗 Μ‚ + 3π‘˜ Μ‚ ) = 17 π‘Ÿ βƒ—.(38𝑖 Μ‚+68𝑗 Μ‚+3π‘˜ Μ‚ ) = 17 Γ— 9 𝒓 βƒ—.(πŸ‘πŸ–π’Š Μ‚+πŸ”πŸ–π’‹ Μ‚+πŸ‘π’Œ Μ‚ ) = 153

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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.