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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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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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.