Vector product - Solving
Last updated at April 16, 2024 by Teachoo
Misc 18 (Introduction) The value of π Μ.(π Μ Γ π Μ) + π Μ. (π Μ Γ π Μ) + π Μ.(π Μ Γ π Μ) (A) 0 (B) β1 (C) 1 (D) 3 π Μ Γ π Μ = π Μ π Μ Γ π Μ = π Μ π Μ Γ π Μ = π Μ π Μ Γ π Μ = βπ Μ π Μ Γ π Μ = βπ Μ π Μ Γ π Μ = βπ Μ Misc 18 The value of π Μ.(π Μ Γ π Μ) + π Μ. (π Μ Γ π Μ) + π Μ.(π Μ Γ π Μ) (A) 0 (B) β1 (C) 1 (D) 3 Now, π Μ.(π Μ Γ π Μ) + π Μ. (π Μ Γ π Μ) + π Μ.(π Μ Γ π Μ) = π Μ.(π Μ) + π Μ.(βπ Μ) + π Μ .(π Μ) = π Μ. π Μ β π Μ . π Μ + π Μ. π Μ = 1 β 1 + 1 = 1 So, option C is correct π Μ. π Μ = |π Μ ||π Μ | cos 0 = 1 Γ 1 Γ 1 = 1 Similarly, π Μ . π Μ = π Μ . π Μ = 1