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 = šŸ/šŸ |(š‘Øš‘©) āƒ— Ɨ (š‘Øš‘Ŗ) āƒ— | Finding AB (š‘Øš‘©) āƒ— = (1 āˆ’ 1) š‘– Ģ‚ + (2 āˆ’ 1) š‘— Ģ‚ + (3 āˆ’ 1) š‘˜ Ģ‚ = 0š‘– Ģ‚ + 1š‘— Ģ‚ + 2š‘˜ Ģ‚ Finding AC (š‘Øš‘Ŗ) āƒ— = (2 āˆ’ 1) š‘– Ģ‚ + (3 āˆ’ 1) š‘— Ģ‚ + (1 āˆ’ 1) š‘˜ Ģ‚ = 1š‘– Ģ‚ + 2š‘— Ģ‚ + 0š‘˜ Ģ‚ (š‘Øš‘©) āƒ— Ɨ (š‘Øš‘Ŗ) āƒ— = |ā– 8(š‘– Ģ‚&š‘— Ģ‚&š‘˜ Ģ‚@0&1&2@1&2&0)| = š‘– Ģ‚ [(1Ɨ0)āˆ’(2Ɨ2)] āˆ’ š‘— Ģ‚[(0Ɨ0)āˆ’(1Ɨ2)] + š‘˜ Ģ‚[(0Ɨ2)āˆ’(1Ɨ1)] = āˆ’4š’Š Ģ‚ + 2š’‹ Ģ‚ – 1š’Œ Ģ‚ Magnitude of (š“šµ) āƒ— Ɨ (š“š¶) āƒ— = √((āˆ’4)2+22+(āˆ’1)2) |(š‘Øš‘©) āƒ—" Ɨ " (š‘Øš‘Ŗ) āƒ— | = √(16+4+1) = āˆššŸšŸ Therefore, Area of triangle ABC = 1/2 |(š“šµ) āƒ—" Ɨ " (š“š¶) āƒ— | = 1/2 Ɨ √21 = āˆššŸšŸ/šŸ

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