Last updated at Dec. 16, 2024 by Teachoo
Question 6 Find the equations of the planes that passes through three points. (b) (1, 1, 0), (1, 2, 1), (โ2, 2, โ1) Vector equation of a plane passing through three points with position vectors ๐ โ, ๐ โ, ๐ โ is ("r" โ โ ๐ โ) . [(๐ โโ๐ โ)ร(๐ โ โ๐ โ)] = 0 Now, the plane passes through the points A (1, 1, 0) ๐ โ = 1๐ ฬ + 1๐ ฬ + 0๐ ฬ B (1, 2, 1) ๐ โ = 1๐ ฬ + 2๐ ฬ + 1๐ ฬ C (โ2, 2, โ1) ๐ โ = โ2๐ ฬ + 2๐ ฬ โ 1๐ ฬ (๐ โ โ ๐ โ) = (1๐ ฬ + 2๐ ฬ + 1๐ ฬ) โ (1๐ ฬ + 1๐ ฬ + 0๐ ฬ) = (1 โ 1) ๐ ฬ + (2 โ 1)๐ ฬ + (1 โ 0) ๐ ฬ = 0๐ ฬ + 1๐ ฬ + 1๐ ฬ (๐ โ โ ๐ โ) = (โ2๐ ฬ + 2๐ ฬ + 1๐ ฬ) โ (1๐ ฬ + 1๐ ฬ +0๐ ฬ) = (โ2 โ 1)๐ ฬ + (2 โ 1)๐ ฬ + (โ1 โ 0) ๐ ฬ = โ3๐ ฬ + 1๐ ฬ โ 1๐ ฬ (๐ โ โ ๐ โ) ร (๐ โ โ ๐ โ) = |โ 8(๐ ฬ&๐ ฬ&๐ ฬ@0&1&1@ โ 3&1& โ 1)| = ๐ ฬ [(1รโ1)โ(1ร1)] โ ๐ ฬ [(0รโ1)โ(โ3 ร1)] + (๐ ) ฬ[(0ร1)โ(โ3 ร1)] = ๐ ฬ(โ1 โ 1) โ ๐ ฬ (0 + 3) + ๐ ฬ ( 0 + 3) = โ2๐ ฬ โ 3๐ ฬ + 3๐ ฬ โด Vector equation of plane is [๐ โโ(1๐ ฬ+1๐ ฬ+0๐ ฬ ) ].(โ2๐ ฬโ3๐ ฬ+3๐ ฬ ) = 0 [๐ โโ(๐ ฬ+๐ ฬ ) ].(โ๐๐ ฬโ๐๐ ฬ+๐๐ ฬ ) = ๐ Finding Cartesian equation Put ๐ โ = x๐ ฬ + y๐ ฬ + z๐ ฬ [๐ โโ(๐ ฬ+๐ ฬ ) ].(โ2๐ ฬโ3๐ ฬ+3๐ ฬ ) = 0 [(๐ฅ๐ ฬ+๐ฆ๐ ฬ+๐ง๐ ฬ )โ(๐ ฬ+๐ ฬ)]. (โ2๐ ฬโ3๐ ฬ+3๐ ฬ ) = 0 [(๐ฅโ1) ๐ ฬ +(๐ฆโ1) ๐ ฬ+๐ง๐ ฬ ]. (โ2๐ ฬโ3๐ ฬ+3๐ ฬ ) = 0 โ2(x โ 1) + (โ3)(y โ 1) + 3(z) = 0 โ2x + 2 โ 3y + 3 + 3z = 0 2x + 3y โ 3z = 5 โด Equation of plane in Cartesian form is 2x + 3y โ 3z = 5
Plane
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About the Author
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