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

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

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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo