Scalar product - Defination

Chapter 10 Class 12 Vector Algebra
Concept wise

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Misc 17 Let π β and π β be two unit vectors and ΞΈ is the angle between them. Then π β + π β is a unit vector if (A) ΞΈ = π/4 (B) ΞΈ = π/3 (C) ΞΈ = π/2 (D) ΞΈ = 2π/3 Given π β & π β are unit vectors, So, |π β | = 1 & |π β | = 1 We need to find ΞΈ if π β + π β is a unit vector Assuming π β + π β is a unit vector Magnitude of π β + π β = 1 |π β+π β |=1 |π β+π β |^π=1^2 (π β+π β ).(π β+π β ) = 1 π β. (π β+π β ) + π β. (π β+π β ) = 1 π β . π β + π β . π β + π β.π β + π β.π β = 1 |π β |^π + π β.π β + π β.π β + |π β |^π=1 12 + π β. π β + π β.π β + 12 = 1 2 + π β. π β + π β.π β = 1 2 + π β. π β + π β. π β = 1 2 + 2 π β. π β = 1 2π β. π β = 1 β 2 π β. π β = (β1)/2 |π β ||π β | cosβ‘ ΞΈ = (β1)/2 1 Γ 1 Γ cos ΞΈ = (β1)/2 cos ΞΈ = (βπ)/π So, ΞΈ = ππ/π Hence, option (D) is correct Rough We know that cos 60Β° = 1/2 & cos is negative in 2nd quadrant, So, ΞΈ = 180 β 60Β° ΞΈ = 120Β° ΞΈ = 120 Γ π/180 ΞΈ = 2π/3