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Example 26 (Supplementary NCERT) Deleted for CBSE Board 2023 Exams You are here

Example 27 (Supplementary NCERT) Deleted for CBSE Board 2023 Exams

Example 28 (Supplementary NCERT) Important Deleted for CBSE Board 2023 Exams

Example 29 (Supplementary NCERT) Important Deleted for CBSE Board 2023 Exams

Example 30 (Supplementary NCERT) Important Deleted for CBSE Board 2023 Exams

Example 31 (Supplementary NCERT) Important Deleted for CBSE Board 2023 Exams

Last updated at March 16, 2023 by Teachoo

Example 26 (Supplementary NCERT) Find π β.(π β Γ π β), if π β = 2π Μ + π Μ + 3π Μ , π β = βπ Μ + 2π Μ + π Μ , π β = 3π Μ + π Μ + 2π ΜGiven, π β = 2π Μ + π Μ + 3π Μ , π β = βπ Μ + 2π Μ + π Μ , π β = 3π Μ + π Μ + 2π Μ π β.(π β Γ π β) = [π β" " π β" " π β ] = |β 8(π&π&π@βπ&π&π@π&π&π)| = 2[(2Γ2)β(1Γ1) ] β 1[(β1Γ2)β(3Γ1) ] + 3[(β1Γ1)β(3Γ2)] = 2 [4β1]β1(β2β3)+3[β1β6] = 2(3) β 1(β5) + 3(β7) = 6 + 5 β 21 = 11 β 21 = β 10 β΄ π β.(π β Γ π β) = β10