Scalar product - Defination

Chapter 10 Class 12 Vector Algebra
Concept wise

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Ex 10.4, 3 If a unit vector π β makes angles π/3 with π Μ, π/4 , with π Μ & an acute angle ΞΈ with π Μ , then find ΞΈ and hence, the components of π β . Let us take a unit vector π β = π₯π Μ + yπ Μ + zπ Μ So, magnitude of π β = |π β | = 1 Angle of π β with π Μ = π/π π β . π Μ = |π β ||π Μ | cos π/3 (xπ Μ + yπ Μ + zπ Μ). π Μ = 1 Γ 1 Γ 1/2 (xπ Μ + yπ Μ + zπ Μ). (1π Μ + 0π Μ + 0π Μ) = 1/2 (x Γ 1) + (y Γ 0) + (z Γ 0) = 1/2 x + 0 + 0 = 1/2 x = π/π Angle of π β with π Μ = π/π π β . π Μ = |π β ||π Μ | cos π/4 (xπ Μ + yπ Μ + zπ Μ). π Μ = 1 Γ 1 Γ 1/β2 (xπ Μ + yπ Μ + zπ Μ). (0π Μ + 1π Μ + 0π Μ) = 1/β2 (x Γ 0) + (y Γ 1) + (z Γ 0) = 1/β2 0 + y + 0 = 1/β2 y = π/βπ Also, Angle of π β with π Μ = ΞΈ π β. π Μ = |π β ||π Μ |Γcosβ‘"ΞΈ" (xπ Μ + yπ Μ + zπ Μ). (0π Μ + .0π Μ + 1π Μ) = 1 Γ 1 Γ cos ΞΈ (x Γ 0) + (y Γ 0) + (z Γ 1) = cosΞΈ 0 + 0 + z = cos ΞΈ z = cos ΞΈ Now, Magnitude of π β = β(π₯^2+π¦2+π§2) 1 = β((1/2)^2+(1/β2)^2+πππ 2"ΞΈ" ) 1 = β(1/4+1/2+πππ 2"ΞΈ" ) 1 = β(3/4+πππ 2"ΞΈ" ) β(3/4+πππ 2"ΞΈ" ) = 1 (β(3/4+πππ 2"ΞΈ" ))^2 = 12 3/4 + πππ 2" ΞΈ" = 1 πππ 2 "ΞΈ" = 1 β 3/4 πππ 2" ΞΈ" = 1/4 cosβ‘"ΞΈ" = Β± β(1/4) cosβ‘"ΞΈ" = Β± 1/2 Since ΞΈ is given an acute angle So, ΞΈ < 90Β° β΄ ΞΈ is in 1st quadrant And, cos ΞΈ is positive in 1st quadrant= So, cos ΞΈ = 1/2 β΄ ΞΈ = 60Β° = π/π Also, z = cos ΞΈ = cos 60Β° = π/π Hence x = 1/2 , y = 1/β2 & z = 1/2 The required vector π β is 1/2 π Μ + 1/β2 π Μ + 1/2 π Μ So, components of π β are π/π , π/βπ & π/π