Last updated at May 6, 2021 by Teachoo
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Ex 10.5, 4 (Supplementary NCERT) Let ๐ โ = ๐ ฬ + ๐ ฬ + ๐ ฬ, ๐ โ = ๐ ฬ and ๐ โ = c1๐ ฬ + c2๐ ฬ + c3๐ ฬ are coplanar (a) If c1 = 1 and c2 = 2, find c3 which makes ๐ โ, ๐ โ, ๐ โ coplanar Given c1 = 1 and c2 = 2 So, our vectors become ๐ โ = ๐ ฬ + ๐ ฬ + ๐ ฬ ๐ โ = ๐ ฬ ๐ โ = c1๐ ฬ + c2๐ ฬ + c3๐ ฬ Three vectors ๐ โ, ๐ โ, ๐ โ are coplanar if [๐ โ" " ๐ โ" " ๐ โ ] = 0 Ex 10.5, 4 (Supplementary NCERT) Let ๐ โ = ๐ ฬ + ๐ ฬ + ๐ ฬ, ๐ โ = ๐ ฬ and ๐ โ = c1๐ ฬ + c2๐ ฬ + c3๐ ฬ are coplanar (b) If c2 = โ1 and c3 = 1, show that no value of c1 can make ๐ โ, ๐ โ, ๐ โ coplanar Given c2 = โ1 and c3 = 1 So, our vectors ๐ โ = ๐ ฬ + ๐ ฬ + ๐ ฬ ๐ โ = ๐ ฬ ๐ โ = c1๐ ฬ + c2๐ ฬ + c3๐ ฬ Three vectors ๐ โ, ๐ โ, ๐ โ are coplanar if [๐ โ" " ๐ โ" " ๐ โ ] = 0 Finding [๐ โ" " ๐ โ" " ๐ โ ] [๐ โ" " ๐ โ" " ๐ โ ] = |โ 8(1&1&1@1&0&0@๐_1&โ1&1)| = 1[(0ร1)โ(0รโ1) ] โ 1[(1ร1)โ(๐_1ร0) ] + 1[(1รโ1)โ(๐_1ร0) ] = 1 [0โ0]โ1[1โ0]+1[โ1โ0] = 0 โ 1 โ 1 = โ2 Thus, [๐ โ" " ๐ โ" " ๐ โ ] โ 0 for any value of c1 So, we can write that ๐ โ,๐ โ,๐ โ are not coplanar for any value of c1
Ex 10.5 (Supplementary NCERT)
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Ex 10.5, 3 (Supplementary NCERT) Deleted for CBSE Board 2022 Exams
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Ex 10.5 (Supplementary NCERT)
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