Ex 10.5 (Supplementary NCERT)
Ex 10.5, 2 (Supplementary NCERT) Deleted for CBSE Board 2023 Exams
Ex 10.5, 3 (Supplementary NCERT) Deleted for CBSE Board 2023 Exams
Ex 10.5, 4 (Supplementary NCERT) Important Deleted for CBSE Board 2023 Exams
Ex 10.5, 5 (Supplementary NCERT) Deleted for CBSE Board 2023 Exams
Ex 10.5, 6 (Supplementary NCERT) Important Deleted for CBSE Board 2023 Exams
Ex 10.5 , 7 (Supplementary NCERT) Important Deleted for CBSE Board 2023 Exams
Ex 10.5 (Supplementary NCERT)
Last updated at May 6, 2021 by Teachoo
Ex 10.5, 1 (Supplementary NCERT) Find [π β" " π β" " π β ] if π β = π Μ β 2π Μ + 3π Μ , π β = 2π Μ β 3π Μ + π Μ , π β = 3π Μ + π Μ β 2π ΜGiven, π β = π Μ β 2π Μ + 3π Μ , π β = 2π Μ β 3π Μ + π Μ , π β = 3π Μ + π Μ β 2π Μ Now, π β.(π β Γ π β) = [π β" " π β" " π β ] = |β 8(π&βπ&π@π&βπ&π@π&π&βπ)| = 1[(β3Γβ2)β(1Γ1) ] β (β2)[(2Γβ2)β(3Γ1) ] + 3[(2Γ1)β(3Γβ3)] = 1 [6β1]+2(β4β3)+3[2+9] = 1(5) + 2(β7) + 3(11) = 5 β 14 + 33 = 38 β 14 = 24 β΄ [π β" " π β" " π β ] = 24