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

  1. Chapter 10 Class 12 Vector Algebra
  2. Serial order wise

Transcript

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 = ๐Ÿ/๐Ÿ |(๐‘จ๐‘ฉ) โƒ— ร— (๐‘จ๐‘ช) โƒ— | Finding AB (๐‘จ๐‘ฉ) โƒ— = (1 โˆ’ 1) ๐‘– ฬ‚ + (2 โˆ’ 1) ๐‘— ฬ‚ + (3 โˆ’ 1) ๐‘˜ ฬ‚ = 0๐‘– ฬ‚ + 1๐‘— ฬ‚ + 2๐‘˜ ฬ‚ Finding AC (๐‘จ๐‘ช) โƒ— = (2 โˆ’ 1) ๐‘– ฬ‚ + (3 โˆ’ 1) ๐‘— ฬ‚ + (1 โˆ’ 1) ๐‘˜ ฬ‚ = 1๐‘– ฬ‚ + 2๐‘— ฬ‚ + 0๐‘˜ ฬ‚ (๐‘จ๐‘ฉ) โƒ— ร— (๐‘จ๐‘ช) โƒ— = |โ– 8(๐‘– ฬ‚&๐‘— ฬ‚&๐‘˜ ฬ‚@0&1&2@1&2&0)| = ๐‘– ฬ‚ [(1ร—0)โˆ’(2ร—2)] โˆ’ ๐‘— ฬ‚[(0ร—0)โˆ’(1ร—2)] + ๐‘˜ ฬ‚[(0ร—2)โˆ’(1ร—1)] = โˆ’4๐’Š ฬ‚ + 2๐’‹ ฬ‚ โ€“ 1๐’Œ ฬ‚ Magnitude of (๐ด๐ต) โƒ— ร— (๐ด๐ถ) โƒ— = โˆš((โˆ’4)2+22+(โˆ’1)2) |(๐‘จ๐‘ฉ) โƒ—" ร— " (๐‘จ๐‘ช) โƒ— | = โˆš(16+4+1) = โˆš๐Ÿ๐Ÿ Therefore, Area of triangle ABC = 1/2 |(๐ด๐ต) โƒ—" ร— " (๐ด๐ถ) โƒ— | = 1/2 ร— โˆš21 = โˆš๐Ÿ๐Ÿ/๐Ÿ

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.