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Ex 10.3, 2 - Find angle between vectors i - 2j + 3k, 3i - 2j + k

Ex 10.3, 2 - Chapter 10 Class 12 Vector Algebra - Part 2
Ex 10.3, 2 - Chapter 10 Class 12 Vector Algebra - Part 3

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Ex 10.3, 2 Find the angle between the vectors 𝑖 Μ‚ βˆ’ 2𝑗 Μ‚ + 3π‘˜ Μ‚ and 3𝑖 Μ‚ - 2𝑗 Μ‚ + π‘˜ Μ‚Let π‘Ž βƒ— = 𝑖 Μ‚ βˆ’ 2𝑗 Μ‚ + 3π‘˜ Μ‚ = 1𝑖 Μ‚ βˆ’ 2𝑗 Μ‚ + 3π‘˜ Μ‚ and 𝑏 βƒ— = 3𝑖 Μ‚ βˆ’ 2𝑗 Μ‚ + π‘˜ Μ‚ = 3𝑖 Μ‚ βˆ’ 2𝑗 Μ‚ + 1π‘˜ Μ‚ We know that π‘Ž βƒ— . 𝑏 βƒ— = |π‘Ž βƒ— ||𝑏 βƒ— | cos ΞΈ ; ΞΈ is the angle between π‘Ž βƒ— & 𝑏 βƒ— Now, 𝒂 βƒ—. 𝒃 βƒ— = (1𝑖 Μ‚ βˆ’ 2𝑗 Μ‚ + 3π‘˜ Μ‚). (3𝑖 Μ‚ βˆ’ 2𝑗 Μ‚ + 1π‘˜ Μ‚) = 1.3 + (βˆ’2).(βˆ’2) + 3.1 = 3 + 4 + 3 = 10 Magnitude of π‘Ž βƒ— = √(12+(βˆ’2)2+32) |π‘Ž βƒ— |= √(1+4+9) = √14 Magnitude of 𝑏 βƒ— = √(32+(βˆ’2)2+12) |𝑏 βƒ— |= √(9+4+1) = √14 Now, π‘Ž βƒ— . 𝑏 βƒ— = |π‘Ž βƒ— ||𝑏 βƒ— | cos ΞΈ 10 = √14 Γ— √14 x cos ΞΈ 10 = 14 Γ— cos ΞΈ cos ΞΈ = 10/14 ΞΈ = cos-1(πŸ“/πŸ•) Thus, the angle between π‘Ž βƒ— and 𝑏 βƒ— is cos-1(5/7)

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