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Ex 10.4, 10 - Find area of parallelogram whose adjacent sides are

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

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Ex 10.4, 10 Find the area of the parallelogram whose adjacent sides are determined by the vectors π‘Ž βƒ— = 𝑖 Μ‚ βˆ’ 𝑗 Μ‚ + 3π‘˜ Μ‚ and b = 2𝑖 Μ‚ βˆ’ 7𝑗 Μ‚ + π‘˜ Μ‚ . π‘Ž βƒ— = 𝑖 Μ‚ βˆ’ 𝑗 Μ‚ + 3π‘˜ Μ‚ = 1𝑖 Μ‚ βˆ’ 1𝑗 Μ‚ + 3k Μ‚ 𝑏 βƒ— = 2𝑖 Μ‚ βˆ’ 7𝑗 Μ‚ + π‘˜ Μ‚ = 2𝑖 Μ‚ βˆ’ 7𝑗 Μ‚ + 1k Μ‚ Area of parallelogram ABCD = |π‘Ž βƒ—" Γ— " 𝑏 βƒ— | 𝒂 βƒ— Γ— 𝒃 βƒ— = |β– 8(𝑖 Μ‚&𝑗 Μ‚&π‘˜ Μ‚@1&βˆ’1&3@2&βˆ’7&1)| = 𝑖 Μ‚ (βˆ’1 Γ— 1 βˆ’ (βˆ’7) Γ— 3) βˆ’ 𝑗 Μ‚ (1 Γ— 1 βˆ’ 2 Γ— 3) + π‘˜ Μ‚ (1 Γ— βˆ’7 βˆ’ 2 Γ— βˆ’1) = 𝑖 Μ‚ (βˆ’1βˆ’(βˆ’21)) βˆ’ 𝑗 Μ‚ (1 βˆ’ 6) + π‘˜ Μ‚ (βˆ’7 βˆ’(βˆ’2)) = 𝑖 Μ‚ (βˆ’1 + 21) βˆ’ 𝑗 Μ‚ (βˆ’5) + π‘˜ Μ‚ (βˆ’7 + 2) = 20 π’Š Μ‚ + 5𝒋 Μ‚ βˆ’ 5π’Œ Μ‚ Magnitude of π‘Ž βƒ— Γ— 𝑏 βƒ— = √(202+52+(βˆ’5)2) |π‘Ž βƒ—" Γ— " 𝑏 βƒ— | = √(400+25+25) = √450 = √(25Γ—9Γ—2) = 5 Γ— 3 Γ— √2 = 15 √2 Therefore, the area of parallelogram is 15√𝟐 .

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