Plane

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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

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


Transcript

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

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Davneet Singh

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.