Scalar product - Solving

Chapter 10 Class 12 Vector Algebra
Concept wise

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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 + (β1 + 3) π Μ + (β3 + (β5)) π Μ = 6π Μ + 2π Μ β 8π Μ (π β β π β) = (5 β 1) π Μ + (β1 β 3) π Μ + (β3 β (β5)) 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 β4) + (β8 Γ 2) = 24 β 8 β16 = 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)