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Misc 6 - Find a vector of magnitude 5 units, parallel to

Misc 6 - Chapter 10 Class 12 Vector Algebra - Part 2
Misc 6 - Chapter 10 Class 12 Vector Algebra - Part 3

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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‚.Given π‘Ž βƒ— = 2𝑖 Μ‚ + 3𝑗 Μ‚ βˆ’ π‘˜ Μ‚(, 𝑏) βƒ— = 𝑖 Μ‚ βˆ’ 2𝑗 Μ‚ + π‘˜ Μ‚ Resultant of 𝒂 βƒ— & 𝒃 βƒ— = 𝒂 βƒ— + 𝒃 βƒ— (𝒂 βƒ— + 𝒃 βƒ—) = (2 + 1)𝑖 Μ‚ + (3 βˆ’ 2)𝑗 Μ‚ + (βˆ’1 + 1)π‘˜ Μ‚ = 3π’Š Μ‚ + 1𝒋 Μ‚ + 0π’Œ Μ‚ Let 𝒄 βƒ— = (𝒂 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‚ + 5/√10 𝑗 Μ‚ + 0π‘˜ Μ‚ = 15/√10 𝑖 Μ‚ + 5/√10 𝑗 Μ‚ Rationalizing = 15/√10 Γ— √10/√10 𝑖 Μ‚ + 5/√10 "Γ— " √10/√10 𝑗 Μ‚ = (15√10)/10 𝑖 Μ‚ + (5√10)/10 𝑗 Μ‚ = (πŸ‘βˆšπŸπŸŽ)/𝟐 π’Š Μ‚ + √𝟏𝟎/𝟐 𝒋 Μ‚ Hence the required vector is (πŸ‘βˆšπŸπŸŽ)/𝟐 π’Š Μ‚ + √𝟏𝟎/𝟐 𝒋 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.