Last updated at Dec. 16, 2024 by Teachoo
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 |(๐ด๐ต) โ ร (๐ด๐ถ) โ | A (1, 1, 2) B (2, 3, 5) (๐ด๐ต) โ = (2 โ 1) ๐ ฬ + (3 โ 1) ๐ ฬ + (5 โ 2) ๐ ฬ = 1๐ ฬ + 2๐ ฬ + 3๐ ฬ A (1, 1, 2) C (1, 5, 5) (๐ด๐ถ) โ = (1 โ 1) ๐ ฬ + (5 โ 1) ๐ ฬ + (5 โ 2) ๐ ฬ = 0๐ ฬ + 4๐ ฬ + 3๐ ฬ (๐จ๐ฉ) โ ร (๐จ๐ช) โ = |โ 8(๐ ฬ&๐ ฬ&๐ ฬ@1&2&3@0&4&3)| = ๐ ฬ (2ร3โ4ร3 )โ๐ ฬ (1ร3โ0ร3 )+๐ ฬ (1ร4โ0ร2 ) = ๐ ฬ (6โ12)โ๐ ฬ(3โ0) + ๐ ฬ (4โ0) = โ6๐ ฬ โ 3๐ ฬ + 4๐ ฬ Magnitude of (๐ด๐ต) โ ร (๐ด๐ถ) โ = โ((โ6)2+(โ3)2+42) |(๐ด๐ต) โ" ร " (๐ด๐ถ) โ | = โ(36+9+16) = โ61 Area of triangle ABC = 1/2 |(๐ด๐ต) โ" ร " (๐ด๐ถ) โ | = 1/2 ร โ61 = โ๐๐/๐
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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