Scalar product - Defination

Chapter 10 Class 12 Vector Algebra
Concept wise

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Ex 10.3, 10 If π β = 2π Μ + 2π Μ + 3π Μ, π β = βπ Μ + 2π Μ + π Μ and π β = 3π Μ + π Μ are such that π β +ππ β is perpendicular to π β , then find the value of π.π β = 2π Μ + 2π Μ + 3π Μ π β = βπ Μ + 2π Μ + π Μ = β1π Μ + 2π Μ + 1π Μ π β = 3π Μ + π Μ = 3π Μ + 1π Μ + 0π Μ Now, (π β + ππ β) = (2π Μ + 2π Μ + 3π Μ) + π (-1π Μ + 2π Μ + 1π Μ) = 2π Μ + 2π Μ + 3π Μ β ππ Μ + 2ππ Μ + ππ Μ = (2 β π) π Μ + (2 + 2π) π Μ + (3 + π) π Μ Since (π β + ππ β) is perpendicular to π β (π β + ππ β). π β = 0 [(2βπ) π Μ+(2+2π) π Μ+(3+π)π Μ ] . (3π Μ + 1π Μ + 0π Μ) = 0 (2 β π).3 + (2 + 2π).1 + (3 + π ).0 = 0 3.2 β 3π + 2 + 2π + 0 = 0 6 β 3π + 2 + 2π = 0 8 β π = 0 π = 8 β΄ π = 8 (Dot product of perpendicular vectors is 0)