Scalar product - Defination

Chapter 10 Class 12 Vector Algebra
Concept wise

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Example 30 If with reference to the right handed system of mutually perpendicular unit vectors π Μ, π Μ and π Μ, "Ξ±" β = 3π Μ β π Μ, "Ξ²" β = 2π Μ + π Μ β 3π Μ, then express "Ξ²" Μ in the form "Ξ²" β = "Ξ²" β1 + "Ξ²" β2, where "Ξ²" β1 is parallel to "Ξ±" β and "Ξ²" β2 is perpendicular to "Ξ±" β.Given πΌ β = 3π Μ β π Μ = 3π Μ β π Μ + 0π Μ "Ξ²" β = 2π Μ + π Μ β 3π Μ = 2π Μ + 1π Μ β 3π Μ To show: "Ξ²" β = "Ξ²" β1 + "Ξ²" β2 Given, "Ξ²" β1 is parallel to πΌ β & "Ξ²" β2 is perpendicular to πΌ β Let "Ξ²" β1 = ππΆ β , π being a scalar. "Ξ²" β1 = π (3π Μ β 1π Μ + 0π Μ) = 3π π Μ β ππ Μ + 0π Μ Now, "Ξ²" β2 = "Ξ²" β β "Ξ²" β1 = ["2" π Μ" + 1" π Μ" β 3" π Μ ] β ["3" ππ Μ" β π" π Μ" + 0" π Μ ] = 2π Μ + 1π Μ β 3π Μ β 3ππ Μ + ππ Μ + 0π Μ = (2 β 3π) π Μ + (1 + π) π Μ β 3π Μ Also, since "Ξ²" β2 is perpendicular to πΌ β "Ξ²" β2 . πΆ β = 0 ["(2 β 3π) " π Μ" + (1 + π) " π Μ" β 3" π Μ ]. (3π Μ β 1π Μ + 0π Μ) = 0 (2 β "3π") Γ 3 + (1 + π) Γ β1 + (β3) Γ 0 = 0 6 β 9"π" β 1 β π = 0 5 β 10π = 0 π = 5/10 π = π/π Putting value of π in "Ξ²" β1 and "Ξ²" β2 , "Ξ²" β1 = 3ππ Μ β ππ Μ + 0π Μ = 3. 1/2 π Μ β 1/2 π Μ + 0 π Μ = π/π π Μ β π/π π Μ "Ξ²" β2 = (2 β 3π) π Μ + (1 + π) π Μ β 3π Μ = ("2 β 3. " 1/2) π Μ + (1+1/2) π Μ β 3π Μ = π/π π Μ + π/π π Μβ 3π Μ "Ξ²" β2 = (2 β 3π) π Μ + (1 + π) π Μ β 3π Μ = ("2 β 3. " 1/2) π Μ + (1+1/2) π Μ β 3π Μ = π/π π Μ + π/π π Μβ 3π Μ Thus, "Ξ²" β1 + "Ξ²" β2 = (π/π " " π Μ" β " π/π " " π Μ )+(π/π " " π Μ" + " π/π " " π Μβ" 3" π Μ ) = 2π Μ + π Μ β 3π Μ = "Ξ²" β Hence proved