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Example 30 Important

Example 26 (Supplementary NCERT) Deleted for CBSE Board 2022 Exams

Example 27 (Supplementary NCERT) Deleted for CBSE Board 2022 Exams

Example 28 (Supplementary NCERT) Important Deleted for CBSE Board 2022 Exams

Example 29 (Supplementary NCERT) Important Deleted for CBSE Board 2022 Exams

Example 30 (Supplementary NCERT) Important Deleted for CBSE Board 2022 Exams You are here

Example 31 (Supplementary NCERT) Important Deleted for CBSE Board 2022 Exams

Last updated at May 6, 2021 by Teachoo

Example 30 (Supplementary NCERT) Prove that [β 8(π β" + " π β&π β+π β&π β+π β )] = 2 [π β , π β, π β] Solving LHS [β 8(π β" + " π β&π β+π β&π β+π β )] = (π β" + " π β). ["(" π β+π β") " Γ " (" π β+π β)] = (π β" + " π β). ["(" π βΓπ β") + (" π βΓπ β)+"(" π βΓπ β") + (" π βΓπ β)] = "(" π β+π β")." [(π βΓπ β") + (" π βΓπ β ) "+ 0 + (" π β" Γ " π β") " ] = π β. (π β Γ π β) + π β.(π β Γ π β) + π β. (π β Γ π β) + π β.(π β Γ π β) + π β.(π β Γ π β) + π β.(π β Γ π β) = [π β", " π β", " π β ] + [π β", " π β", " π β ] + [π β", " π β", " π β ] + [π β", " π β", " π β ] + [π β", " π β", " π β ] + [π β", " π β", " π β ] π β Γ π β = |π β ||π β | sin 0 π Μ = 0 [π β", " π β", " π β ] = [π β", " π β", " π β ] = π β. (π β Γ π β) As (π β Γ π β) = 0 β = π β . 0 β = 0 Using Prop: [π β", " π β", " π β ] = 0 [π β", " π β", " π β ] = 0 [π β", " π β", " π β ] = 0 [π β", " π β", " π β ] = 0 = [π β", " π β", " π β ] + 0 + 0 + 0 + 0 + [π β", " π β", " π β ] = [π β", " π β", " π β ] + [π β", " π β", " π β ] = [π β", " π β", " π β ] + [π β", " π β", " π β ] = 2[π β", " π β", " π β ] = R.H.S Hence proved