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

Transcript

Misc 8(Introduction) Show that the points A(1, โ€“ 2, โ€“ 8), B (5, 0, โ€“ 2) and C (11, 3, 7) are collinear, and find the ratio in which B divides AC. (1) Three points collinear i.e. AB + BC = AC (2) Three vectors. i.e. |(๐ด๐ต) โƒ— | + |(๐ต๐ถ) โƒ— | = |(๐ด๐ถ) โƒ— | Misc 8 Show that the points A(1, โ€“ 2, โ€“ 8), B (5, 0, โ€“ 2) and C (11, 3, 7) are collinear, and find the ratio in which B divides AC. 3 points A, B, C are collinear if i.e. |(๐‘จ๐‘ฉ) โƒ— | + |(๐‘ฉ๐‘ช) โƒ— | = |(๐‘จ๐‘ช) โƒ— | A (1, -2, โˆ’8) B (5, 0,โˆ’2) C (11, 3, 7) (๐ด๐ต) โƒ— = (5 โˆ’ 1) ๐‘– ฬ‚ + (0 โˆ’ (โˆ’2)) ๐‘— ฬ‚ + (โˆ’2โˆ’(โˆ’8)) ๐‘˜ ฬ‚ = 4๐‘– ฬ‚ + 2๐‘— ฬ‚ + 6๐‘˜ ฬ‚ (๐ต๐ถ) โƒ— = (11 โˆ’ 5) ๐‘– ฬ‚ + (3 โˆ’ 0) ๐‘— ฬ‚ + (7โˆ’(โˆ’2)) ๐‘˜ ฬ‚ = 6๐‘– ฬ‚ + 3๐‘— ฬ‚ + 9๐‘˜ ฬ‚ (๐ด๐ถ) โƒ— = (11 โˆ’ 1) ๐‘– ฬ‚ + (3 โˆ’ (โˆ’2)) ๐‘— ฬ‚ + (7โˆ’(โˆ’8)) ๐‘˜ ฬ‚ = 10๐‘– ฬ‚ + 5๐‘— ฬ‚ + 15๐‘˜ ฬ‚ Magnitude of (๐ด๐ต) โƒ— = โˆš(42+22+62) (๐ด๐ต) โƒ— = โˆš(16+4+36) = โˆš56 = โˆš(4ร—14 ) = 2โˆš14 Magnitude of (๐ต๐ถ) โƒ— = โˆš(62+32+92) |(๐ต๐ถ) โƒ— |= โˆš(36+9+81) = โˆš126 = โˆš(9ร—14 ) = 3โˆš14 Magnitude of (๐ด๐ถ) โƒ— = โˆš(102+52+152) |(๐ด๐ถ) โƒ— |= โˆš(100+25+225)= โˆš350 = โˆš(25ร—14 ) = 5โˆš14 |(๐ด๐ต) โƒ— | + |(๐ต๐ถ) โƒ— | = 2โˆš(14 ) + 3โˆš(14 ) = 5โˆš(14 ) = |(๐ด๐ถ) โƒ— | โˆด A, B and C are collinear. Finding the ratio in which B divides AC Let B divide AC in the ratio k : 1 (๐‘‚๐ด) โƒ— = 1๐‘– ฬ‚ โˆ’ 2๐‘— ฬ‚ โˆ’ 8๐‘˜ ฬ‚ (๐‘‚๐ต) โƒ— = 5๐‘– ฬ‚ + 0๐‘— ฬ‚ โˆ’ 2๐‘˜ ฬ‚ and (๐‘‚๐ถ) โƒ— = 11๐‘– ฬ‚ + 3๐‘— ฬ‚ + 7๐‘˜ ฬ‚ Position vector of ๐ต = (๐‘˜(๐‘‚๐ถ) โƒ— + 1.(๐‘‚๐ด) โƒ—)/(๐‘˜ + 1) (๐‘‚๐ต) โƒ— = (๐‘˜(11๐‘– ฬ‚ + 3๐‘— ฬ‚ + 7๐‘˜ ฬ‚ ) + 1(1๐‘– ฬ‚ โˆ’ 2๐‘— ฬ‚ โˆ’ 8๐‘˜ ฬ‚ ))/(๐‘˜ + 1) 5๐‘– ฬ‚ + 0๐‘— ฬ‚ โˆ’ 2๐‘˜ ฬ‚ = (11๐‘˜๐‘– ฬ‚ + 3๐‘˜๐‘— ฬ‚ + 7๐‘˜ ๐‘˜ ฬ‚ + ๐‘– ฬ‚ โˆ’ 2๐‘— ฬ‚ โˆ’ 8๐‘˜ ฬ‚)/(๐‘˜ + 1) 5๐‘– ฬ‚ + 0๐‘— ฬ‚ โˆ’ 2๐‘˜ ฬ‚ = ((11๐‘˜ + 1) ๐‘– ฬ‚ + (3๐‘˜ โˆ’ 2) ๐‘— ฬ‚ + (7๐‘˜ โˆ’ 8) ๐‘˜ ฬ‚)/(๐‘˜ + 1) โˆด 5๐‘– ฬ‚ + 0๐‘— ฬ‚ โˆ’ 2๐‘˜ ฬ‚ = ((11๐‘˜ + 1) )/((๐‘˜ + 1)) ๐‘– ฬ‚ + ((3๐‘˜ โˆ’ 2) )/((๐‘˜ + 1)) ๐‘— ฬ‚ + ((7๐‘˜ โˆ’ 8) )/((๐‘˜ + 1)) ๐‘˜ ฬ‚ Since the two vectors are equal, corresponding components are also equal. So, (11๐‘˜ + 1)/(๐‘˜ + 1) = 5 11k + 1 = 5k + 5 11k โ€“ 5k = 5 โˆ’ 1 6k = 4 k = 4/6 = 2/3 Thus, B divides AC in the ratio 2/3 : 1 or 2 : 3

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.