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  1. Chapter 10 Class 12 Vector Algebra
  2. Serial order wise

Transcript

Misc 6 Find a vector of magnitude 5 units, and parallel to the resultant of the vectors π‘Ž βƒ— = 2𝑖 Μ‚ + 3𝑗 Μ‚ βˆ’ π‘˜ Μ‚ and 𝑏 βƒ— = 𝑖 Μ‚ βˆ’ 2𝑗 Μ‚ + π‘˜ Μ‚. π‘Ž βƒ— = 2𝑖 Μ‚ + 3𝑗 Μ‚ βˆ’ π‘˜ Μ‚ 𝑏 βƒ— = 𝑖 Μ‚ βˆ’ 2𝑗 Μ‚ + π‘˜ Μ‚ (π‘Ž βƒ— + 𝑏 βƒ—) = (2 + 1)𝑖 Μ‚ + (3 βˆ’ 2)𝑗 Μ‚ + (βˆ’1 + 1)π‘˜ Μ‚ = 3𝑖 Μ‚ + 1𝑗 Μ‚ + 0π‘˜ Μ‚ Let 𝑐 βƒ— = (π‘Ž βƒ— + 𝑏 βƒ—) ∴ 𝑐 βƒ— = 3𝑖 Μ‚ + 1𝑗 Μ‚ + 0π‘˜ Μ‚ Magnitude of 𝑐 βƒ— = √(32+12+02) |𝑐 βƒ— | = √(9+1) = √10 Unit vector in direction of 𝑐 βƒ— = 1/|𝑐 βƒ— | Γ— 𝑐 βƒ— 𝑐 Μ‚ = 1/√10 Γ— [3𝑖 Μ‚+1𝑗 Μ‚+0π‘˜ Μ‚ ] 𝑐 Μ‚ = πŸ‘/√𝟏𝟎 π’Š Μ‚ + 𝟏/√𝟏𝟎 𝒋 Μ‚ + 0π’Œ Μ‚ Vector with magnitude 1 = 3/√10 𝑖 Μ‚ + 1/√10 𝑗 Μ‚ + 0π‘˜ Μ‚ Vector with magnitude 5 = 5 Γ— [3/√10 " " 𝑖 Μ‚" + " 1/√10 𝑗 Μ‚" + 0" π‘˜ Μ‚ ] = 15/√10 𝑖 Μ‚ + 2/√10 𝑗 Μ‚ + 0π‘˜ Μ‚ = 15/√10 𝑖 Μ‚ + 2/√10 𝑗 Μ‚ Rationalizing = 15/√10 Γ— √10/√10 𝑖 Μ‚ + 2/√10 "Γ— " √10/√10 𝑗 Μ‚ = (15√10)/10 𝑖 Μ‚ + (2√10)/10 𝑗 Μ‚ = (3√10)/2 𝑖 Μ‚ + √10/5 𝑗 Μ‚ Hence the required vector is (πŸ‘βˆšπŸπŸŽ)/𝟐 π’Š Μ‚ + √𝟏𝟎/πŸ“ 𝒋 Μ‚

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.