Maths Crash Course - Live lectures + all videos + Real time Doubt solving!

Examples

Example 1
Important

Example 2 Important

Example 3 Important

Example 4

Example 5

Example 6 Important

Example 7

Example 8 Important

Example 9

Example 10

Example 11 (i) Important

Example 11 (ii) Important

Example 12 Important

Example 13

Example 14 Important

Example 15

Example 16 Important

Example 17

Example 18 Important

Example 19

Example 20

Example 21 Important

Example 22

Example 23 Important

Example 24

Example 25 Important

Example 26

Example 27 Important

Example 28 Important

Example 29 Important

Example 30 Important

Example 26 (Supplementary NCERT) Deleted for CBSE Board 2023 Exams

Example 27 (Supplementary NCERT) Deleted for CBSE Board 2023 Exams

Example 28 (Supplementary NCERT) Important Deleted for CBSE Board 2023 Exams

Example 29 (Supplementary NCERT) Important Deleted for CBSE Board 2023 Exams

Example 30 (Supplementary NCERT) Important Deleted for CBSE Board 2023 Exams You are here

Example 31 (Supplementary NCERT) Important Deleted for CBSE Board 2023 Exams

Last updated at May 6, 2021 by Teachoo

Maths Crash Course - Live lectures + all videos + Real time Doubt solving!

Example 30 (Supplementary NCERT) Prove that [β 8(π β" + " π β&π β+π β&π β+π β )] = 2 [π β , π β, π β] Solving LHS [β 8(π β" + " π β&π β+π β&π β+π β )] = (π β" + " π β). ["(" π β+π β") " Γ " (" π β+π β)] = (π β" + " π β). ["(" π βΓπ β") + (" π βΓπ β)+"(" π βΓπ β") + (" π βΓπ β)] = "(" π β+π β")." [(π βΓπ β") + (" π βΓπ β ) "+ 0 + (" 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) + π β.(π β Γ 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", " π β ] π β Γ π β = |π β ||π β | sin 0 π Μ = 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 ] = 0 [π β", " π β", " π β ] = 0 [π β", " π β", " π β ] = 0 [π β", " π β", " π β ] = 0 = [π β", " π β", " π β ] + 0 + 0 + 0 + 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 ] = 2[π β", " π β", " π β ] = R.H.S Hence proved