Vector product - Solving

Chapter 10 Class 12 Vector Algebra
Concept wise

This video is only available for Teachoo black users

Introducing your new favourite teacher - Teachoo Black, at only βΉ83 per month

### Transcript

Ex 10.4, 4 Show that (π β β π β) Γ (π β + π β) = 2(π β Γ π β) Solving L.H.S (π β β π β) Γ (π β + π β) = π β Γ (π β + π β) β π β Γ (π β + π β) = π β Γ π β + π β Γ π β β π β Γ π β β π β Γ π β = π β Γ π β + π β Γ π β β (β π β Γ π β) β π β Γ π β = π β Γ π β + π β Γ π β + π β Γ π β β π β Γ π β = π β Γ π β + 2(π β Γ π β) β π β Γ π β π β Γ π β = |π β ||π β | sin ΞΈ π Μ = |π β |2 sin 0 π Μ = 0 So, π β Γ π β = 0 Similarly, π β Γ π β = 0 = 0 + 2(π β Γ π β) β 0 = 2 (π β Γ π β) = RHS Since LHS = RHS, Hence proved.