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Example 26 (Supplementary NCERT) Deleted for CBSE Board 2023 Exams

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Example 30 (Supplementary NCERT) Important Deleted for CBSE Board 2023 Exams

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Last updated at March 30, 2023 by Teachoo

Example 24 Find the area of a triangle having the points A(1, 1, 1), B(1, 2, 3) and C(2, 3, 1) as its vertices. Given A (1, 1, 1) , B (1, 2, 3) ,C (2, 3, 1) Area of triangle ABC = π/π |(π¨π©) β Γ (π¨πͺ) β | Finding AB (π¨π©) β = (1 β 1) π Μ + (2 β 1) π Μ + (3 β 1) π Μ = 0π Μ + 1π Μ + 2π Μ Finding AC (π¨πͺ) β = (2 β 1) π Μ + (3 β 1) π Μ + (1 β 1) π Μ = 1π Μ + 2π Μ + 0π Μ (π¨π©) β Γ (π¨πͺ) β = |β 8(π Μ&π Μ&π Μ@0&1&[email protected]&2&0)| = π Μ [(1Γ0)β(2Γ2)] β π Μ[(0Γ0)β(1Γ2)] + π Μ[(0Γ2)β(1Γ1)] = β4π Μ + 2π Μ β 1π Μ Magnitude of (π΄π΅) β Γ (π΄πΆ) β = β((β4)2+22+(β1)2) |(π¨π©) β" Γ " (π¨πͺ) β | = β(16+4+1) = βππ Therefore, Area of triangle ABC = 1/2 |(π΄π΅) β" Γ " (π΄πΆ) β | = 1/2 Γ β21 = βππ/π