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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 You are here
Ex 10.5 (Supplementary NCERT)
Last updated at March 23, 2023 by Teachoo
Ex 10.5, 7 (Supplementary NCERT) Show that vectors π β, π β and π β are coplanar if π β + π β, π β + π and π β + π β are coplanarGiven π β + π β, π β + π and π β + π β are coplanar [β 8(π β" + " π β&π β+π β&π β+π β )] = 0 We need to prove π β, π β and π β are coplanar i.e. [β 8(π β&π β&π β )] = 0 Now, [β 8(π β" + " π β&π β+π β&π β+π β )] = 0 (π β" + " π β). ["(" π β+π β") " Γ " (" π β+π β)] = 0 (π β" + " π β). ["(" π β Γ π β") + (" π β Γ π β)+"(" π β Γ π β") + (" π β Γ π β)] = 0 "(" π β+π β")." [(π β Γ π β") + (" π β Γ π β ) "+ 0 + (" π β" Γ " π β") " ] = 0 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) = 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 ] + [π β", " π β", " π β ] = 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 ] = π [π β", " π β", " π β ] = π [π β", " π β", " π β ] = π [π β", " π β", " π β ] = π [π β", " π β", " π β ] + 0 + 0 + 0 + 0 + [π β", " π β", " π β ] = 0 [π β", " π β", " π β ] + [π β", " π β", " π β ] = 0 [π β", " π β", " π β ] + [π β", " π β", " π β ] = 0 2[π β", " π β", " π β ] = 0 [π β", " π β", " π β ] = 0 Since [π β", " π β", " π β ] = 0, π β, π β and π β are coplanar As [π β", " π β", " π β ] = [π β", " π β", " π β ]= [π β", " π β", " π β ]