Miscellaneous
Last updated at August 11, 2026 by Teachoo
Transcript
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