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Ex 11.3
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Last updated at March 16, 2023 by Teachoo
Ex 11.3, 6 Find the equations of the planes that passes through three points. (b) (1, 1, 0), (1, 2, 1), (β2, 2, β1) Vector equation of a plane passing through three points with position vectors π β, π β, π β is ("r" β β π β) . [(π ββπ β)Γ(π β βπ β)] = 0 Now, the plane passes through the points A (1, 1, 0) π β = 1π Μ + 1π Μ + 0π Μ B (1, 2, 1) π β = 1π Μ + 2π Μ + 1π Μ C (β2, 2, β1) π β = β2π Μ + 2π Μ β 1π Μ (π β β π β) = (1π Μ + 2π Μ + 1π Μ) β (1π Μ + 1π Μ + 0π Μ) = (1 β 1) π Μ + (2 β 1)π Μ + (1 β 0) π Μ = 0π Μ + 1π Μ + 1π Μ (π β β π β) = (β2π Μ + 2π Μ + 1π Μ) β (1π Μ + 1π Μ +0π Μ) = (β2 β 1)π Μ + (2 β 1)π Μ + (β1 β 0) π Μ = β3π Μ + 1π Μ β 1π Μ (π β β π β) Γ (π β β π β) = |β 8(π Μ&π Μ&π Μ@0&1&[email protected] β 3&1& β 1)| = π Μ [(1Γβ1)β(1Γ1)] β π Μ [(0Γβ1)β(β3 Γ1)] + (π ) Μ[(0Γ1)β(β3 Γ1)] = π Μ(β1 β 1) β π Μ (0 + 3) + π Μ ( 0 + 3) = β2π Μ β 3π Μ + 3π Μ β΄ Vector equation of plane is [π ββ(1π Μ+1π Μ+0π Μ ) ].(β2π Μβ3π Μ+3π Μ ) = 0 [π ββ(π Μ+π Μ ) ].(βππ Μβππ Μ+ππ Μ ) = π Finding Cartesian equation Put π β = xπ Μ + yπ Μ + zπ Μ [π ββ(π Μ+π Μ ) ].(β2π Μβ3π Μ+3π Μ ) = 0 [(π₯π Μ+π¦π Μ+π§π Μ )β(π Μ+π Μ)]. (β2π Μβ3π Μ+3π Μ ) = 0 [(π₯β1) π Μ +(π¦β1) π Μ+π§π Μ ]. (β2π Μβ3π Μ+3π Μ ) = 0 β2(x β 1) + (β3)(y β 1) + 3(z) = 0 β2x + 2 β 3y + 3 + 3z = 0 2x + 3y β 3z = 5 β΄ Equation of plane in Cartesian form is 2x + 3y β 3z = 5