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Example 26 (Supplementary NCERT) Deleted for CBSE Board 2022 Exams

Example 27 (Supplementary NCERT) Deleted for CBSE Board 2022 Exams

Example 28 (Supplementary NCERT) Important Deleted for CBSE Board 2022 Exams

Example 29 (Supplementary NCERT) Important Deleted for CBSE Board 2022 Exams

Example 30 (Supplementary NCERT) Important Deleted for CBSE Board 2022 Exams

Example 31 (Supplementary NCERT) Important Deleted for CBSE Board 2022 Exams

Last updated at May 6, 2021 by Teachoo

Example 15 If π β = 5π Μ β π Μ β 3π Μ and π β = π Μ + 3π Μ β 5π Μ , then show that the vectors π β + π β and π β β π β are perpendicular. Two vectors π β and π β are perpendicular if their scalar product is zero, i.e. π β . π β = 0 Finding (π β + π β) and (π β β π β) (π β + π β) = (5 + 1) π Μ + (β1 + 3) π Μ + (β3 + (β5)) π Μ = 6π Μ + 2π Μ β 8π Μ (π β β π β) = (5 β 1) π Μ + (β1 β 3) π Μ + (β3 β (β5)) π Μ = 4π Μ β 4π Μ + 2π Μ We have to show that (π β + π β) and (π β β π β) are perpendicular to each other. So, we need to show (π β + π β) . (π β β π β) = 0 Solving LHS (π β + π β) . (π β β π β) = (6π Μ + 2π Μ β 8π Μ) . (4π Μ β 4π Μ + 2π Μ) = (6 Γ 4) + (2 Γ β4) + (β8 Γ 2) = 24 β 8 β16 = 0 Since (π β + π β) . (π β β π β) = 0 Hence, (π β + π β) is perpendicular to (π β β π β)