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Example 26 (Supplementary NCERT) Deleted for CBSE Board 2022 Exams

Example 27 (Supplementary NCERT) Deleted for CBSE Board 2022 Exams

Example 28 (Supplementary NCERT) Important Deleted for CBSE Board 2022 Exams

Example 29 (Supplementary NCERT) Important Deleted for CBSE Board 2022 Exams

Example 30 (Supplementary NCERT) Important Deleted for CBSE Board 2022 Exams

Example 31 (Supplementary NCERT) Important Deleted for CBSE Board 2022 Exams

Last updated at May 6, 2021 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&2@1&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