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Ex 11.3

Ex 11.3, 1 (a)
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Chapter 11 Class 12 Three Dimensional Geometry

Serial order wise

Last updated at Aug. 24, 2021 by Teachoo

Ex 11.3, 6 (Introduction) Find the equations of the planes that passes through three points. (a) (1, 1, β 1), (6, 4, β 5), (β 4, β 2, 3) 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 ββπ β)] = 0 Ex 11.3, 6 Find the equations of the planes that passes through three points. (a) (1, 1, β1), (6, 4, β5), (β4, β2, 3) 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 ββπ β)] = 0 Now, the plane passes through the points (π β β π β) = (6π Μ + 4π Μ β 5π Μ) β (1π Μ + 1π Μ β 1π Μ) = (6 β1)π Μ + (4 β 1)π Μ + (β5 β (β1)) π Μ = 5π Μ + 3π Μ β 4π Μ A (1, 1, β1) π β = 1π Μ + 1π Μ β 1π Μ B (6, 4, β5) π β = 6π Μ + 4π Μ β 5π Μ C ( β4, β2, 3) π β = β4π Μ β 2π Μ + 3π Μ (π β β π β) = (β4π Μ β 2π Μ + 3π Μ) β (1π Μ + 1π Μ β 1π Μ) = (β4 β 1)π Μ +(β2 β 1)π Μ + (3 β (β1)) π Μ = β5π Μ β 3π Μ + 4π Μ (π β β π β) Γ (π β β π β) = |β 8(π Μ&π Μ&π Μ@5&3&β4@β5&β3&4)| = β |β 8(π Μ&π Μ&π Μ@5&3&β4@ 5& 3&β4)| = π β This implies, the three points are collinear. Using property: Since the two rows of the determinant are same, the value of determinant is zero. β΄ Vector equation of plane is [π ββ(π Μ+π Μ βπ Μ )] . 0 β = 0 Since, the above equation is satisfied for all values of π β, Therefore, there will be infinite planes passing through the given 3 collinear points.