Ex 10.5 (Supplementary NCERT)
Ex 10.5, 2 (Supplementary NCERT) Deleted for CBSE Board 2022 Exams
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Ex 10.5, 5 (Supplementary NCERT) Deleted for CBSE Board 2022 Exams
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Ex 10.5 (Supplementary NCERT)
Last updated at May 6, 2021 by Teachoo
Ex 10.5, 4 (Supplementary NCERT) Let π β = π Μ + π Μ + π Μ, π β = π Μ and π β = c1π Μ + c2π Μ + c3π Μ are coplanar (a) If c1 = 1 and c2 = 2, find c3 which makes π β, π β, π β coplanar Given c1 = 1 and c2 = 2 So, our vectors become π β = π Μ + π Μ + π Μ π β = π Μ π β = c1π Μ + c2π Μ + c3π Μ Three vectors π β, π β, π β are coplanar if [π β" " π β" " π β ] = 0 Ex 10.5, 4 (Supplementary NCERT) Let π β = π Μ + π Μ + π Μ, π β = π Μ and π β = c1π Μ + c2π Μ + c3π Μ are coplanar (b) If c2 = β1 and c3 = 1, show that no value of c1 can make π β, π β, π β coplanar Given c2 = β1 and c3 = 1 So, our vectors π β = π Μ + π Μ + π Μ π β = π Μ π β = c1π Μ + c2π Μ + c3π Μ Three vectors π β, π β, π β are coplanar if [π β" " π β" " π β ] = 0 Finding [π β" " π β" " π β ] [π β" " π β" " π β ] = |β 8(1&1&1@1&0&0@π_1&β1&1)| = 1[(0Γ1)β(0Γβ1) ] β 1[(1Γ1)β(π_1Γ0) ] + 1[(1Γβ1)β(π_1Γ0) ] = 1 [0β0]β1[1β0]+1[β1β0] = 0 β 1 β 1 = β2 Thus, [π β" " π β" " π β ] β 0 for any value of c1 So, we can write that π β,π β,π β are not coplanar for any value of c1