Vector product - Defination

Chapter 10 Class 12 Vector Algebra
Concept wise

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Ex 10.4, 5 Find Ξ» and ΞΌ if (2π Μ + 6π Μ + 27π Μ) Γ (π Μ + πj Μ + ΞΌπ Μ) = 0 β Let π β = 2π Μ + 6π Μ + 27π Μ & π β = 1π Μ + πj Μ + ΞΌπ Μ Given, π β Γ π β = π β Γ π β = |β 8(π Μ&π Μ&π Μ@β(2@1)&β(6@"π" )&β(27@"ΞΌ" ))| = π Μ [(6Γ"ΞΌ" )β("π" Γ27)] β π Μ [(2Γ"ΞΌ" )β(1Γ27) ] + π Μ[(2Γ"π" )β(1Γ6)] = π Μ [6"ΞΌ" β27"π" ] β π Μ [2"ΞΌ" β27 ] + π Μ[(2"π" β6)] β΄ π β Γ π β = [6"ΞΌ" β27"π" ] π Μ β (2"ΞΌ"β27) π Μ + (2"π" β 6) π Μ Also, π β Γ π β = 0 β [π"ΞΌ" βππ"π" ] π Μ β (π"ΞΌ"βππ) π Μ + (2"π" β 6) π Μ = 0π Μ + 0π Μ + 0π Μ Comparing components Therefore, "π" = 3 and "ΞΌ" = ππ/π π Μ 6"ΞΌ" β 27"π" = 0 6"ΞΌ" β 27"π" "ΞΌ" = 27/6 "π" π Μ β (2"ΞΌ" β 27) = 0 2"ΞΌ" β 27 = 0 2"ΞΌ" = 27 "ΞΌ" = 27/2 π Μ (2"π" β 6) = 0 2"π" = 6 "π" = 3