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Chapter 11 Class 12 Three Dimensional Geometry
Ex 11.1, 2
Example, 6 Important
Example, 7
Example 10 Important
Ex 11.2, 5 Important
Ex 11.2, 9 (i) Important
Ex 11.2, 10 Important
Ex 11.2, 12 Important
Ex 11.2, 13 Important
Ex 11.2, 15 Important
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Misc 3 Important
Misc 4 Important
Question 10 Important Deleted for CBSE Board 2024 Exams You are here
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Misc 5 Important
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Chapter 11 Class 12 Three Dimensional Geometry
Last updated at May 29, 2023 by Teachoo
Question 10 Find the vector equation of the plane passing through the intersection of the planes 𝑟 ⃗ . (𝑖 ̂ + 𝑗 ̂ + 𝑘 ̂) = 6 and 𝑟 ⃗ . (2𝑖 ̂ + 3𝑗 ̂ + 4𝑘 ̂) = − 5, and the point (1, 1, 1).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 ⃗.(𝒊 ̂ + 𝒋 ̂ + 𝒌 ̂) = 6 Comparing with 𝑟 ⃗.(𝑛1) ⃗ = d1, (𝒏𝟏) ⃗ = 𝒊 ̂ + 𝒋 ̂ + 𝒌 ̂ & d1 = 6 𝒓 ⃗.(2𝒊 ̂ + 3𝒋 ̂ + 4𝒌 ̂) = −5 –𝑟 ⃗.(2𝑖 ̂ + 3𝑗 ̂ + 4𝑘 ̂) = 5 𝑟 ⃗ .(− 2𝑖 ̂ − 3𝑗 ̂ − 4𝑘 ̂) = 5 Comparing with 𝑟 ⃗.(𝑛2) ⃗ = d2 (𝒏𝟐) ⃗ = − 2𝒊 ̂ − 3𝒋 ̂ − 4𝒌 ̂ & d2 = 5 Equation of plane is 𝑟 ⃗. [(𝑖 ̂+𝑗 ̂+𝑘 ̂ )+"𝜆" (−2𝑖 ̂−3𝑗 ̂−4𝑘 ̂)] = 6 + 𝜆5 𝒓 ⃗. [(𝒊 ̂" " +𝒋 ̂" " +𝒌 ̂ )−"𝜆" (𝟐𝒊 ̂+𝟑𝒋 ̂+𝟒𝒌 ̂)] = 6 + 5𝜆 Now to find 𝜆 , put 𝒓 ⃗ = x𝒊 ̂ + y𝒋 ̂ + z𝒌 ̂ (x𝑖 ̂ + y𝑗 ̂ + z𝑘 ̂). [(𝑖 ̂+𝑗 ̂+𝑘 ̂ )−"𝜆" (2𝑖 ̂+3𝑗 ̂+4𝑘 ̂)] = 5𝜆 + 6 (x𝑖 ̂ + y𝑗 ̂ + z𝑘 ̂).(𝑖 ̂+𝑗 ̂+𝑘 ̂ ) − 𝜆 (x𝑖 ̂ + y𝑗 ̂ + z𝑘 ̂).(2𝑖 ̂+3𝑗 ̂+4𝑘 ̂) = 5𝜆 + 6 (x × 1) + (y × 1) + (z × 1) − 𝜆[(𝑥×2)+(𝑦×3)+(𝑧×4)] = 5𝜆 + 6 x + y + z − 𝜆[2𝑥+3𝑦+4𝑧] = 5𝜆 + 6 x + y + z − 2𝜆𝑥 − 3𝜆y − 4𝜆z = 5𝜆 + 6 (1 − 2𝜆)x + (1 − 3𝜆)y + (1 − 4𝜆) z = 5𝜆 + 6 Since the plane passes through (1, 1, 1), Putting (1, 1, 1) in (2) (1 − 2𝜆)x + (1 − 3𝜆)y + (1 − 4𝜆) z = 5𝜆 + 6 (1 −2𝜆) × 1 + (1 − 3𝜆) × 1 + (1 − 4𝜆) × 1 = 5𝜆 + 6 1 −2𝜆 + 1 − 3𝜆 + 1 − 4𝜆= 5𝜆 + 6 3 − 9𝜆 = 5𝜆 + 6 −14𝜆 = 3 ∴ 𝜆 = (−𝟑)/𝟏𝟒 Putting value of 𝜆 in (1), 𝑟 ⃗. [(𝑖 ̂" " +" " 𝑗 ̂" " +" " 𝑘 ̂ )−(( −3)/14)(2𝑖 ̂+3𝑗 ̂+"4" 𝑘 ̂)]= 6 + 5 × ( −3)/14 𝑟 ⃗. [(𝑖 ̂+𝑗 ̂+" " 𝑘 ̂ )+3/14(2𝑖 ̂+3𝑗 ̂+"4" 𝑘 ̂)]= 6 − 15/14 𝑟 ⃗. [𝑖 ̂+𝑗 ̂" " +𝑘 ̂+ 6/14 𝑖 ̂+9/14 𝑗 ̂+12/14 𝑘 ̂ ]= 69/14 𝑟 ⃗. [(1+6/14) 𝑖 ̂ +(1+9/14) 𝑗 ̂+(1+12/14) 𝑘 ̂ ]= 69/14 𝑟 ⃗. [20/14 𝑖 ̂ + 23/14 𝑗 ̂ + 26/14 𝑘 ̂ ]= 69/14 𝑟 ⃗. [1/14(20𝑖 ̂+23𝑗 ̂+26𝑘 ̂)]= 69/14 1/14 𝑟 ⃗. (20𝑖 ̂ + 23𝑗 ̂ + 26𝑘 ̂) = 69/14 𝑟 ⃗. (20𝑖 ̂ + 23𝑗 ̂ + 26𝑘 ̂) = 69 Therefore, the vector equation of the required plane is 𝒓 ⃗.(𝟐𝟎𝒊 ̂ + 𝟐𝟑𝒋 ̂ + 𝟐𝟔𝒌 ̂) = 𝟔𝟗