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Example 24 - Find area of a triangle having A (1, 1, 1), B (1, 2, 3)

Example 24 - Chapter 10 Class 12 Vector Algebra - Part 2

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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 = 𝟏/𝟐 |(𝑨𝑩) βƒ— Γ— (𝑨π‘ͺ) βƒ— | Finding AB (𝑨𝑩) βƒ— = (1 βˆ’ 1) 𝑖 Μ‚ + (2 βˆ’ 1) 𝑗 Μ‚ + (3 βˆ’ 1) π‘˜ Μ‚ = 0𝑖 Μ‚ + 1𝑗 Μ‚ + 2π‘˜ Μ‚ Finding AC (𝑨π‘ͺ) βƒ— = (2 βˆ’ 1) 𝑖 Μ‚ + (3 βˆ’ 1) 𝑗 Μ‚ + (1 βˆ’ 1) π‘˜ Μ‚ = 1𝑖 Μ‚ + 2𝑗 Μ‚ + 0π‘˜ Μ‚ (𝑨𝑩) βƒ— Γ— (𝑨π‘ͺ) βƒ— = |β– 8(𝑖 Μ‚&𝑗 Μ‚&π‘˜ Μ‚@0&1&2@1&2&0)| = 𝑖 Μ‚ [(1Γ—0)βˆ’(2Γ—2)] βˆ’ 𝑗 Μ‚[(0Γ—0)βˆ’(1Γ—2)] + π‘˜ Μ‚[(0Γ—2)βˆ’(1Γ—1)] = βˆ’4π’Š Μ‚ + 2𝒋 Μ‚ – 1π’Œ Μ‚ Magnitude of (𝐴𝐡) βƒ— Γ— (𝐴𝐢) βƒ— = √((βˆ’4)2+22+(βˆ’1)2) |(𝑨𝑩) βƒ—" Γ— " (𝑨π‘ͺ) βƒ— | = √(16+4+1) = √𝟐𝟏 Therefore, Area of triangle ABC = 1/2 |(𝐴𝐡) βƒ—" Γ— " (𝐴𝐢) βƒ— | = 1/2 Γ— √21 = √𝟐𝟏/𝟐

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