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Supplementary examples and questions from CBSE
Supplementary Example 2 Important Deleted for CBSE Board 2023 Exams
Supplementary Example 3 Deleted for CBSE Board 2023 Exams
Supplementary Example 4 Deleted for CBSE Board 2023 Exams
Supplementary Example 5 Important Deleted for CBSE Board 2023 Exams
Supplementary Example 6 Deleted for CBSE Board 2023 Exams
Supplementary Exercise Q1 Deleted for CBSE Board 2023 Exams
Supplementary Exercise Q2 Deleted for CBSE Board 2023 Exams
Supplementary Exercise Q3 Deleted for CBSE Board 2023 Exams
Supplementary Exercise Q4 Important Deleted for CBSE Board 2023 Exams
Supplementary Exercise Q5 Deleted for CBSE Board 2023 Exams You are here
Supplementary Exercise Q6 Important Deleted for CBSE Board 2023 Exams
Supplementary Exercise Q7 Deleted for CBSE Board 2023 Exams
Supplementary Exercise Q8 Deleted for CBSE Board 2023 Exams
Supplementary Exercise Q9 Important Deleted for CBSE Board 2023 Exams
Supplementary Exercise Q10 Deleted for CBSE Board 2023 Exams
Supplementary examples and questions from CBSE
Last updated at Aug. 24, 2021 by Teachoo
Supplementary Exercise Q5 Find the value of ๐ if the following vectors are co-planar (i) ๐ โ = 2๐ ฬ โ ๐ ฬ + ๐ ฬ, ๐ โ = ๐ ฬ + 2๐ ฬ โ 3๐ ฬ, and ๐ โ = 3๐ ฬ โ ๐๐ ฬ + 5๐ ฬ Three vectors ๐ โ, ๐ โ, ๐ โ are coplanar if [๐ โ" " ๐ โ" " ๐ โ ] = 0 Given, ๐ โ = 2๐ ฬ โ ๐ ฬ + ๐ ฬ ๐ โ = ๐ ฬ + 2๐ ฬ โ 3๐ ฬ ๐ โ = 3๐ ฬ โ ฮป ๐ ฬ + 5๐ ฬ [๐ โ" " ๐ โ" " ๐ โ ] = |โ 8(2&โ1&1@1&2&โ3@3&โ๐&5)| 0 = 2[(2ร5)โ(โฮปรโ3) ] โ (โ1) [(1ร5)โ(3รโ3) ] + 1[(1รโฮป)โ(3ร2) ] 0 = 2[10โ3ฮป] + [5+9 ] + [โฮปโ6] 0 = 20 โ 6ฮป + 14 โ ฮป +โ 6 0 = โ7ฮป + 28 7ฮป = 28 ฮป = 28/7 ๐ = 4 Therefore, ๐ โ,๐,๐ โ are coplanar if ฮป = 4 Supplementary Exercise Q5 Find the value of ๐ if the following vectors are co-planar (ii) ๐ โ = 2๐ ฬ + ๐ ฬ + ๐ ฬ, ๐ โ = 2๐ ฬ โ ๐๐ ฬ + ๐ ฬ, and ๐ โ = 5๐ ฬ โ ๐ ฬ โ 3๐ ฬ Three vectors ๐ โ, ๐ โ, ๐ โ are coplanar if [๐ โ" " ๐ โ" " ๐ โ ] = 0 Given, ๐ โ = 2๐ ฬ + ๐ ฬ + ๐ ฬ ๐ โ = 2๐ ฬ โ ฮป๐ ฬ + ๐ ฬ ๐ โ = 5๐ ฬ + ๐ ฬ โ 3๐ ฬ [๐ โ" " ๐ โ" " ๐ โ ] = |โ 8(2&1&1@2&โ๐&1@5&1&โ3)| 0 = 2[(โฮปรโ3)โ(1ร1)] โ 1[(2รโ3)โ(5ร1)] + 1[(2ร1)โ(5รโฮป) ] 0 = 2[3ฮปโ1] โ 1[โ6โ5] + 1[2+5ฮป] 0 = 6ฮป โ 2 + 11 + 2 + 5ฮป 0 = 11ฮป + 11 โ11ฮป = 11 ฮป = 11/(โ11) ๐ = โ1 Therefore, ๐ โ,๐,๐ โ are coplanar if ฮป = โ1