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Misc 7 - Find unit vector parallel to vector 2a - b + 3c

Misc 7 - Chapter 10 Class 12 Vector Algebra - Part 2

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Misc 7 If π‘Ž βƒ— = 𝑖 Μ‚ + 𝑗 Μ‚ + π‘˜ Μ‚, 𝑏 βƒ— = 2𝑖 Μ‚ βˆ’π‘— Μ‚ + 3π‘˜ Μ‚ and 𝑐 βƒ— = 𝑖 Μ‚ βˆ’ 2𝑗 Μ‚ + π‘˜ Μ‚ , find a unit vector parallel to the vector 2π‘Ž βƒ— – 𝑏 βƒ— + 3𝑐 βƒ— . Given π‘Ž βƒ— = 𝑖 Μ‚ + 𝑗 Μ‚ + π‘˜ Μ‚ 𝑏 βƒ— = 2𝑖 Μ‚ + 𝑗 Μ‚ + 3π‘˜ Μ‚ 𝑐 βƒ— = 𝑖 Μ‚ βˆ’ 2𝑗 Μ‚ + π‘˜ Μ‚ Let 𝒓 βƒ— = 2𝒂 βƒ— βˆ’ 𝒃 βƒ— + 3𝒄 βƒ— = 2(𝑖 Μ‚ + 𝑗 Μ‚ + π‘˜ Μ‚) βˆ’ (2𝑖 Μ‚ βˆ’ 𝑗 Μ‚ + 3π‘˜ Μ‚) + 3(𝑖 Μ‚ βˆ’2𝑗 Μ‚ + π‘˜ Μ‚) = 2𝑖 Μ‚ + 2𝑗 Μ‚ + 2π‘˜ Μ‚ βˆ’ 2𝑖 Μ‚ + 1𝑗 Μ‚ βˆ’ 3π‘˜ Μ‚ + 3𝑖 Μ‚ βˆ’ 6𝑗 Μ‚ + 3π‘˜ Μ‚ = (2 βˆ’ 2 + 3) 𝑖 Μ‚ + (2 + 1 βˆ’ 6) 𝑗 Μ‚ + (2 βˆ’ 3 + 3) π‘˜ Μ‚ = 3π’Š Μ‚ – 3𝒋 Μ‚ + 2π’Œ Μ‚ ∴ 𝒓 βƒ— = 3π’Š Μ‚ – 3𝒋 Μ‚ + 2π’Œ Μ‚ Magnitude of π‘Ÿ βƒ— = √(32+(βˆ’3)2+22) |𝒓 βƒ— | = √(9+9+4) = √𝟐𝟐 Unit vector in the direction of π‘Ÿ βƒ— = 𝟏/|𝒓 βƒ— | x 𝒓 βƒ— = 1/√22 Γ— [3𝑖 Μ‚ βˆ’3𝑗 Μ‚+2π‘˜ Μ‚ ] = 3/√22 𝑖 Μ‚ – 3/√22 𝑗 Μ‚ + 2/√22 π‘˜ Μ‚ Hence the required vector is πŸ‘/√𝟐𝟐 π’Š Μ‚ – πŸ‘/√𝟐𝟐 𝒋 Μ‚ + 𝟐/√𝟐𝟐 π’Œ Μ‚

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.