Last updated at Dec. 16, 2024 by Teachoo
Misc 13 The scalar product of the vector ๐ ฬ + ๐ ฬ + ๐ ฬ with a unit vector along the sum of vectors 2๐ ฬ + 4๐ ฬ โ 5๐ ฬ and ฮป๐ ฬ + 2๐ ฬ + 3๐ ฬ is equal to one. Find the value of ฮป. Let ๐ โ = ๐ ฬ + ๐ ฬ + ๐ ฬ ๐ โ = 2๐ ฬ + 4๐ ฬ โ 5๐ ฬ ๐ โ = ๐ ๐ ฬ + 2๐ ฬ + 3๐ ฬ (๐ โ + ๐ โ) = (2 + ๐) ๐ ฬ + (4 + 2) ๐ ฬ + (โ5 + 3) ๐ ฬ = (2 + ๐) ๐ ฬ + 6๐ ฬ โ 2๐ ฬ Let ๐ ฬ be unit vector along (๐ โ + ๐ โ) ๐ ฬ = 1/(๐๐๐๐๐๐ก๐ข๐๐ ๐๐ (๐ โ" + " ๐ โ)) ร (๐ โ + ๐ โ) ๐ ฬ = 1/โ((2 + ๐)^2 + 6^2 + (โ2)^2 ) ร ((2 + ๐) ๐ ฬ + 6๐ ฬ โ 2๐ ฬ) ๐ ฬ = 1/โ(2^2 + ๐^2 + 4๐ + 36 + 4) ร ((2 + ๐) ๐ ฬ + 6๐ ฬ โ 2๐ ฬ) ๐ ฬ = ๐/โ(๐^๐ + ๐๐ +๐๐) ร ((2 + ๐) ๐ ฬ + 6๐ ฬ โ 2๐ ฬ) Given, ๐ โ. (๐ ฬ) = 1 (1๐ ฬ + 1๐ ฬ + 1๐ ฬ). (1/โ(๐^2 + 4๐ +44) " ร ((2 + ๐) " ๐ ฬ" + 6" ๐ ฬ" โ 2" ๐ ฬ")" ) = 1 1/โ(๐^2 + 4๐ +44) (1๐ ฬ + 1๐ ฬ + 1๐ ฬ).((๐ +2) ๐ ฬ + 6๐ ฬ โ 2๐ ฬ) = 1 (1๐ ฬ + 1๐ ฬ + 1๐ ฬ).((๐ +2) ๐ ฬ + 6๐ ฬ โ 2๐ ฬ) = โ(๐^2 + 4๐ +44) 1.(๐ + 2) + 1.6 + 1.(โ2) = โ(๐^2 + 4๐ +44) ๐ + 2 + 6 โ 2 = โ(๐^2 + 4๐ +44) ๐ + 6 = โ(๐^๐ + ๐๐ +๐๐) Squaring both sides (๐ + 6)2 = (โ(๐^2 + 4๐ +44))^2 ๐2 + 36 + 12๐ = ๐^2 + 4๐ +44 8๐ = 8 ๐ = 8/8 ๐ = 1 So, ๐ = 1
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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