Get live Maths 1-on-1 Classs - Class 6 to 12
Example 22 - Chapter 10 Class 12 Vector Algebra (Term 2)
Last updated at March 30, 2023 by Teachoo
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 You are here
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
Example 31 (Supplementary NCERT) Important Deleted for CBSE Board 2023 Exams
Transcript
Example 22 Find |π β Γ π β|, if π β = 2π Μ + π Μ + 3π Μ and π β = 3π Μ + 5π Μ β 2π Μ Given π β = 2π Μ + π Μ + 3π Μ π β = 3π Μ + 5π Μ β 2π Μ Now, π β Γ π β = |β 8(π Μ&π Μ&π Μ@2&1&[email protected]&5&β2)| = π Μ (1 Γ (-2) β 5 Γ 3) β π Μ (2 Γ (-2) β 3 Γ 3) + π Μ (2 Γ 5 β 3 Γ 1) = π Μ (β2 β15) β π Μ (β4 β9) + π Μ (10 β 3) = β17π Μ + 13π Μ + 7π Μ So, Magnitude of π β Γ π β = β((β17)2+(13)2+(7)2) |π β" Γ " π β | = β(289+169+49) = β507 Thus, |π β" Γ " π β | = βπππ
