Vector product - Defination

Chapter 10 Class 12 Vector Algebra
Concept wise

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Ex 10.4, 7 Let the vectors π β π β, π β be given as π_1 π Μ + π_2 π Μ +π_3 π Μ, π_1 π Μ + π_2 π Μ +π_3 π Μ, π_1 π Μ + π_2 π Μ +π_3 π Μ Then show that π β Γ (π β + π β) =π β Γπ β + π β Γ π β. Let π β = π_1 π Μ + π_2 π Μ +π_3 π Μ π β = π_1 π Μ + π_2 π Μ +π_3 π Μ, π β = π_1 π Μ + π_2 π Μ + π_3 π Μ We need to show : π β Γ (π β + π β) = π β Γ π β + π β Γ π β RHS (π β Γ π β) = |β 8(π Μ&π Μ&π Μ@π1&π2&π3@π1&π2&π3)| = (π2 π3 β π2 π3) π Μ β (π1 π3 β π1 π3) π Μ + (π1 π2 β π1 π2) π Μ (π β Γ π β) = |β 8(π Μ&π Μ&π Μ@π1&π2&π3@π1&π2&π3)| = (π2" " π3 β π2" " π3) π Μ β (π1" " π3 β π1" " π3) π Μ + (π1" " π2 β π1" " π2) π Μ (π β Γ π β) + (π β Γ π β) = [π2" " π3βπ2 π3+π2 π3βπ2π3] π Μ β [π1" " π3βπ1 π3+π1 π3βπ1π3] π Μ + [π1" " π2βπ1 π2+π1 π2βπ1π2] π Μ Since the corresponding components are equal, So, π β Γ (π β + π β) = π β Γ π β + π β Γ π β Hence proved. Ex 10.4, 7 Let the vectors , be given as 1 + 2 + 3 , 1 + 2 + 3 , 1 + 2 + 3 Then show that ( + ) = + . Let = 1 + 2 + 3 = 1 + 2 + 3 , = 1 + 2 + 3 We need to show : ( + ) = + LHS ( + ) = ( 1 + 1) + ( 2 + 2) + ( 3 + 3) ( + ) = 1 ( 1+ 1) 2 ( 2+ 2) 3 ( 3+ 3) = 2 ( 3 + 3) ( 2 + 2) 3 1 ( 3 + 3) ( 1 + 1) 3 + 1 ( 2 + 2) ( 1 + 1) 2 = 2 3+ 2 3 2 3 2 3 1 3 1 3+ 1 3 1 3 + 1 2+ 1 2 1 2 1 2