Find a vector in direction of vector 5i -j + 2k which has magnitude 8

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Ex 10.2, 10 - Chapter 10 Class 12 Vector Algebra - Part 2

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

Transcript

Ex 10.2, 10 Find a vector in the direction of vector 5𝑖 Μ‚ βˆ’ 𝑗 Μ‚ + 2π‘˜ Μ‚ which has magnitude 8 units.π‘Ž βƒ— = 5𝑖 Μ‚ βˆ’ 𝑗 Μ‚ + 2π‘˜ Μ‚ = 5𝑖 Μ‚ βˆ’ 1𝑗 Μ‚ + 2π‘˜ Μ‚ Magnitude of π‘Ž βƒ— = √(52+(βˆ’1)2+22) |π‘Ž βƒ— | = √(25+1+4) = √30 Unit vector in direction of π‘Ž βƒ— = 1/|π‘Ž βƒ— | . π‘Ž βƒ— π‘Ž Μ‚ = 1/√30 . [5𝑖 Μ‚βˆ’1𝑗 Μ‚+2π‘˜ Μ‚ ] π‘Ž Μ‚ = 5/√30 𝑖 Μ‚ βˆ’ 1/√30 𝑗 Μ‚ + 2/√30 π‘˜ Μ‚ Thus, unit vector π‘Ž Μ‚ = 5/√30 𝑖 Μ‚ βˆ’ 1/√30 𝑗 Μ‚ + 2/√30 π‘˜ Μ‚ Vector with magnitude 1 = 5/√30 𝑖 Μ‚ βˆ’ 1/√30 𝑗 Μ‚ + 2/√30 π‘˜ Μ‚ Vector with magnitude 8 = 8 [5/√30 𝑖 Μ‚" βˆ’ " 1/√30 𝑗 Μ‚" + " 2/√30 π‘˜ Μ‚ ] = πŸ’πŸŽ/βˆšπŸ‘πŸŽ π’Š Μ‚ – πŸ–/βˆšπŸ‘πŸŽ 𝒋 Μ‚ + πŸπŸ”/βˆšπŸ‘πŸŽ π’Œ Μ‚ Hence, the required vector is 40/√30 𝑖 Μ‚ – 8/√30 𝑗 Μ‚ + 16/√30 π‘˜ 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 10 years. He provides courses for Maths and Science at Teachoo.