Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

Plane

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Chapter 11 Class 12 Three Dimensional Geometry

Serial order wise

Last updated at May 29, 2023 by Teachoo

Question 10 Find the vector equation of plane passing through the intersection of the planes 𝑟 ⃗ . (2𝑖 ̂ + 2𝑗 ̂ − 3𝑘 ̂) = 7, 𝑟 ⃗ .(2𝑖 ̂ + 5𝑗 ̂ + 3𝑘 ̂) = 9 and through the point (2, 1, 3).The vector equation of a plane passing through the intersection of planes 𝑟 ⃗. (𝑛1) ⃗ = d1 and 𝑟 ⃗. (𝑛2) ⃗ = d2 and also through the point (x1, y1, z1) is 𝒓 ⃗.((𝒏𝟏) ⃗ + 𝜆(𝒏𝟐) ⃗) = d1 + 𝜆d2 Given, the plane passes through The vector equation of a plane passing through the intersection of planes 𝑟 ⃗. (𝑛1) ⃗ = d1 and 𝑟 ⃗. (𝑛2) ⃗ = d2 and also through the point (x1, y1, z1) is 𝒓 ⃗.((𝒏𝟏) ⃗ + 𝜆(𝒏𝟐) ⃗) = d1 + 𝜆d2 Given, the plane passes through 𝒓 ⃗. (2𝒊 ̂ + 2𝒋 ̂ − 3𝒌 ̂) = 7 Comparing with 𝑟 ⃗.(𝑛1) ⃗ = 𝑑1, (𝑛1) ⃗ = 2𝑖 ̂ + 2𝑗 ̂ − 3𝑘 ̂ & d1 = 7 𝒓 ⃗. (2𝒊 ̂ + 5𝒋 ̂ + 3𝒌 ̂) = 9 Comparing with 𝑟 ⃗.(𝑛2) ⃗ = 𝑑2, (𝑛2) ⃗ = 2𝑖 ̂ + 5𝑗 ̂ + 3𝑘 ̂ & d2 = 9 So, equation of the plane is 𝑟 ⃗.["(2" 𝑖 ̂+"2" 𝑗 ̂" " −"3" 𝑘 ̂")" +"𝜆(2" 𝑖 ̂" " + 5𝑗 ̂" " + "3" 𝑘 ̂")" ] = 7 + 𝜆.9 𝑟 ⃗. ["2" 𝑖 ̂" " +" 2" 𝑗 ̂" " − "3" 𝑘 ̂ + 2"𝜆" 𝑖 ̂ + 5"𝜆" 𝑗 ̂ + 3"𝜆" 𝑘 ̂ ] = 7 + 9"𝜆" 𝒓 ⃗. ["(2" +"2𝜆" )𝒊 ̂" " +"(2" +"5𝜆" )𝒋 ̂ +"(−" 𝟑+"3𝜆" )𝒌 ̂ ] = 9"𝜆" + 7 Now, to find 𝜆 , put 𝒓 ⃗ = x𝒊 ̂ + y𝒋 ̂ + z𝒌 ̂ (x𝑖 ̂ + y𝑗 ̂ + z𝑘 ̂).["(2 " + "2𝜆" )𝑖 ̂" " + "(2" +"5𝜆" )𝑗 ̂ +"(−" 3" " +" 3𝜆" )𝑘 ̂ ] = 9𝜆 + 7 x"(2 "+" 2𝜆")" "+ "y (2 "+" 5𝜆")𝑗 ̂ + 𝑧"("−3+"3𝜆")𝑘 ̂ = 9𝜆 + 7 The plane passes through (2, 1, 3) Putting (2, 1, 3) in (2), 2(2 + 2𝜆) + 1(2 + 5𝜆) + 3(−3 + 3𝜆) = 9𝜆 + 7 4 + 4𝜆 + 2 + 5𝜆 + (−9) + 9𝜆 = 9𝜆 + 7 18𝜆 − 9𝜆 = 7 + 3 9𝜆 = 10 ∴ 𝜆 = 𝟏𝟎/𝟗 Putting value of 𝜆 in (1), 𝑟 ⃗. [(2+ 2. 10/9) 𝑖 ̂ + (2+ 5. 10/9) 𝑗 ̂ +("−" 3+ 3. 10/9) 𝑘 ̂ ] = 9.10/9 + 7 𝑟 ⃗. [(2 + 20/9) 𝑖 ̂ +(2 + 50/9) 𝑗 ̂ +("−" 3 + 30/9) 𝑘 ̂ ] = 10 + 7 𝑟 ⃗. [38/9 𝑖 ̂+ 68/9 𝑗 ̂ + 3/9 𝑘 ̂ ] = 17 1/9 𝑟 ⃗. (38𝑖 ̂ + 68𝑗 ̂ + 3𝑘 ̂ ) = 17 𝑟 ⃗.(38𝑖 ̂+68𝑗 ̂+3𝑘 ̂ ) = 17 × 9 𝒓 ⃗.(𝟑𝟖𝒊 ̂+𝟔𝟖𝒋 ̂+𝟑𝒌 ̂ ) = 153