Ex 11.3, 6 - Chapter 11 Class 12 Three Dimensional Geometry - Part 5

Ex 11.3, 6 - Chapter 11 Class 12 Three Dimensional Geometry - Part 6
Ex 11.3, 6 - Chapter 11 Class 12 Three Dimensional Geometry - Part 7

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Question 6 Find the equations of the planes that passes through three points. (b) (1, 1, 0), (1, 2, 1), (–2, 2, –1) Vector equation of a plane passing through three points with position vectors 𝑎 ⃗, 𝑏 ⃗, 𝑐 ⃗ is ("r" ⃗ − 𝒂 ⃗) . [(𝒃 ⃗−𝒂 ⃗)×(𝒄 ⃗ −𝒂 ⃗)] = 0 Now, the plane passes through the points A (1, 1, 0) 𝑎 ⃗ = 1𝑖 ̂ + 1𝑗 ̂ + 0𝑘 ̂ B (1, 2, 1) 𝑏 ⃗ = 1𝑖 ̂ + 2𝑗 ̂ + 1𝑘 ̂ C (−2, 2, −1) 𝑐 ⃗ = −2𝑖 ̂ + 2𝑗 ̂ − 1𝑘 ̂ (𝒃 ⃗ − 𝒂 ⃗) = (1𝑖 ̂ + 2𝑗 ̂ + 1𝑘 ̂) − (1𝑖 ̂ + 1𝑗 ̂ + 0𝑘 ̂) = (1 − 1) 𝑖 ̂ + (2 − 1)𝑗 ̂ + (1 − 0) 𝑘 ̂ = 0𝒊 ̂ + 1𝒋 ̂ + 1𝒌 ̂ (𝒄 ⃗ − 𝒂 ⃗) = (–2𝑖 ̂ + 2𝑗 ̂ + 1𝑘 ̂) − (1𝑖 ̂ + 1𝑗 ̂ +0𝑘 ̂) = (−2 − 1)𝑖 ̂ + (2 − 1)𝑗 ̂ + (−1 − 0) 𝑘 ̂ = −3𝒊 ̂ + 1𝒋 ̂ − 1𝒌 ̂ (𝒃 ⃗ − 𝒂 ⃗) × (𝒄 ⃗ − 𝒂 ⃗) = |■8(𝑖 ̂&𝑗 ̂&𝑘 ̂@0&1&1@ − 3&1& − 1)| = 𝑖 ̂ [(1×−1)−(1×1)] – 𝑗 ̂ [(0×−1)−(−3 ×1)] + (𝑘 ) ̂[(0×1)−(−3 ×1)] = 𝑖 ̂(–1 – 1) – 𝑗 ̂ (0 + 3) + 𝑘 ̂ ( 0 + 3) = –2𝒊 ̂ – 3𝒋 ̂ + 3𝒌 ̂ ∴ Vector equation of plane is [𝑟 ⃗−(1𝑖 ̂+1𝑗 ̂+0𝑘 ̂ ) ].(−2𝑖 ̂−3𝑗 ̂+3𝑘 ̂ ) = 0 [𝒓 ⃗−(𝒊 ̂+𝒋 ̂ ) ].(−𝟐𝒊 ̂−𝟑𝒋 ̂+𝟑𝒌 ̂ ) = 𝟎 Finding Cartesian equation Put 𝒓 ⃗ = x𝒊 ̂ + y𝒋 ̂ + z𝒌 ̂ [𝑟 ⃗−(𝑖 ̂+𝑗 ̂ ) ].(−2𝑖 ̂−3𝑗 ̂+3𝑘 ̂ ) = 0 [(𝑥𝑖 ̂+𝑦𝑗 ̂+𝑧𝑘 ̂ )−(𝑖 ̂+𝑗 ̂)]. (−2𝑖 ̂−3𝑗 ̂+3𝑘 ̂ ) = 0 [(𝑥−1) 𝑖 ̂ +(𝑦−1) 𝑗 ̂+𝑧𝑘 ̂ ]. (−2𝑖 ̂−3𝑗 ̂+3𝑘 ̂ ) = 0 –2(x − 1) + (−3)(y − 1) + 3(z) = 0 –2x + 2 − 3y + 3 + 3z = 0 2x + 3y – 3z = 5 ∴ Equation of plane in Cartesian form is 2x + 3y – 3z = 5

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo