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Example 26 (Supplementary NCERT) Deleted for CBSE Board 2022 Exams You are here
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
Example 31 (Supplementary NCERT) Important Deleted for CBSE Board 2022 Exams
Last updated at May 6, 2021 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
