Vector product - Solving

Chapter 10 Class 12 Vector Algebra
Concept wise

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Ex 10.4, 8 If either π β = 0 β or π β = 0 β, then π β Γ π β = 0 β . Is the converse true? Justify your answer with an example. Converse : If π β Γ π β = 0 β, then either π β = 0 β or π β = 0 β π β Γ π β = |π β ||π β | sin ΞΈ π Μ where, ΞΈ = angle between π β and π β π Μ = unit vector perpendicular to π β π π β Let π β = 1π Μ + 1π Μ + 1π Μ & π β = 2π Μ + 2π Μ + 2π Μ π β Γ π β = |β 8(π Μ&π Μ&π Μ@1&1&1@2&2&2)| = π Μ (1 Γ 2 β 2 Γ 1) β π Μ (1 Γ 2 β2 Γ 1) + π Μ (1 Γ 2 β 2 Γ 1) + π Μ(1 Γ 2 β 2 Γ 1) = π Μ (2 β 2) β π Μ (2 β2) + π Μ (2 β 2) = 0π Μ β 0π Μ + 0π Μ = 0 β Here, π β β  0 β & π ββ  0 β But π β Γ π β = 0 β Therefore, converse is not true.