Ex 10.4, 9 - Find area of triangle A(1, 1, 2), B(2, 3, 5)

Ex 10.4, 9 - Chapter 10 Class 12 Vector Algebra - Part 2

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Ex 10.4, 9 Find the area of the triangle with vertices A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5). A (1, 1, 2) , B (2, 3, 5) , C (1, 5, 5) Area of triangle ABC is 1/2 |(š“šµ) āƒ— Ɨ (š“š¶) āƒ— | A (1, 1, 2) B (2, 3, 5) (š“šµ) āƒ— = (2 āˆ’ 1) š‘– Ģ‚ + (3 āˆ’ 1) š‘— Ģ‚ + (5 āˆ’ 2) š‘˜ Ģ‚ = 1š‘– Ģ‚ + 2š‘— Ģ‚ + 3š‘˜ Ģ‚ A (1, 1, 2) C (1, 5, 5) (š“š¶) āƒ— = (1 āˆ’ 1) š‘– Ģ‚ + (5 āˆ’ 1) š‘— Ģ‚ + (5 āˆ’ 2) š‘˜ Ģ‚ = 0š‘– Ģ‚ + 4š‘— Ģ‚ + 3š‘˜ Ģ‚ (š‘Øš‘©) āƒ— Ɨ (š‘Øš‘Ŗ) āƒ— = |ā– 8(š‘– Ģ‚&š‘— Ģ‚&š‘˜ Ģ‚@1&2&3@0&4&3)| = š‘– Ģ‚ (2Ɨ3āˆ’4Ɨ3 )āˆ’š‘— Ģ‚ (1Ɨ3āˆ’0Ɨ3 )+š‘˜ Ģ‚ (1Ɨ4āˆ’0Ɨ2 ) = š‘– Ģ‚ (6āˆ’12)āˆ’š‘— Ģ‚(3āˆ’0) + š‘˜ Ģ‚ (4āˆ’0) = āˆ’6š’Š Ģ‚ – 3š’‹ Ģ‚ + 4š’Œ Ģ‚ Magnitude of (š“šµ) āƒ— Ɨ (š“š¶) āƒ— = √((āˆ’6)2+(āˆ’3)2+42) |(š“šµ) āƒ—" Ɨ " (š“š¶) āƒ— | = √(36+9+16) = √61 Area of triangle ABC = 1/2 |(š“šµ) āƒ—" Ɨ " (š“š¶) āƒ— | = 1/2 Ɨ √61 = āˆššŸ”šŸ/šŸ

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