Example 26 (Supplementary NCERT)
Last updated at Dec. 5, 2017 by Teachoo
Transcript
Example 26 (Supplementary NCERT) Find πΒ β.(πΒ β Γ πΒ β), if πΒ β = 2πΒ Μ + πΒ Μ + 3πΒ Μ , πΒ β = βπΒ Μ + 2πΒ Μ + πΒ Μ , πΒ β = 3πΒ Μ + πΒ Μ + 2πΒ ΜGiven, πΒ β = 2πΒ Μ + πΒ Μ + 3πΒ Μ , πΒ β = βπΒ Μ + 2πΒ Μ + πΒ Μ , πΒ β = 3πΒ Μ + πΒ Μ + 2πΒ Μ πΒ β.(πΒ β Γ πΒ β) = [πΒ β" " πΒ β" " πΒ β ] = |β 8(2&1&3@β1&2&1@3&1&2)| = 2[(2Γ2)β(1Γ1) ] β 1[(β1Γ2)β(3Γ1) ] + 3[(β1Γ1)β(3Γ2)] = 2 [4β1]β1(β2β3)+3[β1β6] = 2(3) β 1(β5) + 3(β7) = 6 + 5 β 21 = 11 β 21 = β 19 β΄ πΒ β.(πΒ β Γ πΒ β) = β19
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Example 21 Important
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Example 24 Important
Example 25 Important
Example 26
Example 27
Example 28 Important
Example 29 Important
Example 30 Important
Example 26 (Supplementary NCERT) You are here
Example 27 (Supplementary NCERT)
Example 28 (Supplementary NCERT)
Example 29 (Supplementary NCERT)
Example 30 (Supplementary NCERT)
Example 31 (Supplementary NCERT)
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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 7 years. He provides courses for Mathematics and Science from Class 6 to 12. You can learn personally from here https://www.teachoo.com/premium/maths-and-science-classes/.