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Scalar product - Defination
Scalar product - Defination
Last updated at Aug. 23, 2021 by Teachoo
Misc 13 The scalar product of the vector π Μ + π Μ + π Μ with a unit vector along the sum of vectors 2π Μ + 4π Μ β 5π Μ and Ξ»π Μ + 2π Μ + 3π Μ is equal to one. Find the value of Ξ». Let π β = π Μ + π Μ + π Μ π β = 2π Μ + 4π Μ β 5π Μ π β = π π Μ + 2π Μ + 3π Μ (π β + π β) = (2 + π) π Μ + (4 + 2) π Μ + (β5 + 3) π Μ = (2 + π) π Μ + 6π Μ β 2π Μ Let π Μ be unit vector along (π β + π β) π Μ = 1/(ππππππ‘π’ππ ππ (π β" + " π β)) Γ (π β + π β) π Μ = 1/β((2 + π)^2 + 6^2 + (β2)^2 ) Γ ((2 + π) π Μ + 6π Μ β 2π Μ) π Μ = 1/β(2^2 + π^2 + 4π + 36 + 4) Γ ((2 + π) π Μ + 6π Μ β 2π Μ) π Μ = π/β(π^π + ππ +ππ) Γ ((2 + π) π Μ + 6π Μ β 2π Μ) Given, π β. (π Μ) = 1 (1π Μ + 1π Μ + 1π Μ). (1/β(π^2 + 4π +44) " Γ ((2 + π) " π Μ" + 6" π Μ" β 2" π Μ")" ) = 1 1/β(π^2 + 4π +44) (1π Μ + 1π Μ + 1π Μ).((π +2) π Μ + 6π Μ β 2π Μ) = 1 (1π Μ + 1π Μ + 1π Μ).((π +2) π Μ + 6π Μ β 2π Μ) = β(π^2 + 4π +44) 1.(π + 2) + 1.6 + 1.(β2) = β(π^2 + 4π +44) π + 2 + 6 β 2 = β(π^2 + 4π +44) π + 6 = β(π^π + ππ +ππ) Squaring both sides (π + 6)2 = (β(π^2 + 4π +44))^2 π2 + 36 + 12π = π^2 + 4π +44 8π = 8 π = 8/8 π = 1 So, π = 1