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Ex 10.4, 4 - Show that (a - b) x (a + b) = 2(a x b) - Ex 10.4

Ex 10.4, 4 - Chapter 10 Class 12 Vector Algebra - Part 2

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Ex 10.4, 4 Show that (π‘Ž βƒ— βˆ’ 𝑏 βƒ—) Γ— (π‘Ž βƒ— + 𝑏 βƒ—) = 2(π‘Ž βƒ— Γ— 𝑏 βƒ—) Solving L.H.S (π‘Ž βƒ— βˆ’ 𝑏 βƒ—) Γ— (π‘Ž βƒ— + 𝑏 βƒ—) = π‘Ž βƒ— Γ— (π‘Ž βƒ— + 𝑏 βƒ—) βˆ’ 𝑏 βƒ— Γ— (π‘Ž βƒ— + 𝑏 βƒ—) = π‘Ž βƒ— Γ— π‘Ž βƒ— + π‘Ž βƒ— Γ— 𝑏 βƒ— βˆ’ 𝒃 βƒ— Γ— 𝒂 βƒ— βˆ’ 𝑏 βƒ— Γ— 𝑏 βƒ— = π‘Ž βƒ— Γ— π‘Ž βƒ— + π‘Ž βƒ— Γ— 𝑏 βƒ— βˆ’ (βˆ’ 𝒂 βƒ— Γ— 𝒃 βƒ—) βˆ’ 𝑏 βƒ— Γ— 𝑏 βƒ— = π‘Ž βƒ— Γ— π‘Ž βƒ— + π‘Ž βƒ— Γ— 𝑏 βƒ— + π‘Ž βƒ— Γ— 𝑏 βƒ— βˆ’ 𝑏 βƒ— Γ— 𝑏 βƒ— = 𝒂 βƒ— Γ— 𝒂 βƒ— + 2(π‘Ž βƒ— Γ— 𝑏 βƒ—) βˆ’ 𝒃 βƒ— Γ— 𝒃 βƒ— π‘Ž βƒ— Γ— π‘Ž βƒ— = |π‘Ž βƒ— ||π‘Ž βƒ— | sin ΞΈ 𝑛 Μ‚ = |π‘Ž βƒ— |2 sin 0 𝑛 Μ‚ = 0 So, π‘Ž βƒ— Γ— π‘Ž βƒ— = 0 Similarly, 𝑏 βƒ— Γ— 𝑏 βƒ— = 0 = 0 + 2(π‘Ž βƒ— Γ— 𝑏 βƒ—) βˆ’ 0 = 2 (π‘Ž βƒ— Γ— 𝑏 βƒ—) = RHS Since LHS = RHS, Hence proved.

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