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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 |(๐ด๐ต) โƒ— ร— (๐ด๐ถ) โƒ— | A (1, 1, 2) B (2, 3, 5) (๐ด๐ต) โƒ— = (2 โˆ’ 1) ๐‘– ฬ‚ + (3 โˆ’ 1) ๐‘— ฬ‚ + (5 โˆ’ 2) ๐‘˜ ฬ‚ = 1๐‘– ฬ‚ + 2๐‘— ฬ‚ + 3๐‘˜ ฬ‚ A (1, 1, 2) C (1, 5, 5) (๐ด๐ถ) โƒ— = (1 โˆ’ 1) ๐‘– ฬ‚ + (5 โˆ’ 1) ๐‘— ฬ‚ + (5 โˆ’ 2) ๐‘˜ ฬ‚ = 0๐‘– ฬ‚ + 4๐‘— ฬ‚ + 3๐‘˜ ฬ‚ (๐‘จ๐‘ฉ) โƒ— ร— (๐‘จ๐‘ช) โƒ— = |โ– 8(๐‘– ฬ‚&๐‘— ฬ‚&๐‘˜ ฬ‚@1&2&3@0&4&3)| = ๐‘– ฬ‚ (2ร—3โˆ’4ร—3 )โˆ’๐‘— ฬ‚ (1ร—3โˆ’0ร—3 )+๐‘˜ ฬ‚ (1ร—4โˆ’0ร—2 ) = ๐‘– ฬ‚ (6โˆ’12)โˆ’๐‘— ฬ‚(3โˆ’0) + ๐‘˜ ฬ‚ (4โˆ’0) = โˆ’6๐’Š ฬ‚ โ€“ 3๐’‹ ฬ‚ + 4๐’Œ ฬ‚ Magnitude of (๐ด๐ต) โƒ— ร— (๐ด๐ถ) โƒ— = โˆš((โˆ’6)2+(โˆ’3)2+42) |(๐ด๐ต) โƒ—" ร— " (๐ด๐ถ) โƒ— | = โˆš(36+9+16) = โˆš61 Area of triangle ABC = 1/2 |(๐ด๐ต) โƒ—" ร— " (๐ด๐ถ) โƒ— | = 1/2 ร— โˆš61 = โˆš๐Ÿ”๐Ÿ/๐Ÿ

  1. Chapter 10 Class 12 Vector Algebra
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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo