Last updated at April 22, 2021 by

Transcript

Ex 10.2, 10 Find a vector in the direction of vector 5π Μ β π Μ + 2π Μ which has magnitude 8 units.π β = 5π Μ β π Μ + 2π Μ = 5π Μ β 1π Μ + 2π Μ Magnitude of π β = β(52+(β1)2+22) |π β | = β(25+1+4) = β30 Unit vector in direction of π β = 1/|π β | . π β π Μ = 1/β30 . [5π Μβ1π Μ+2π Μ ] π Μ = 5/β30 π Μ β 1/β30 π Μ + 2/β30 π Μ Thus, unit vector π Μ = 5/β30 π Μ β 1/β30 π Μ + 2/β30 π Μ Vector with magnitude 1 = 5/β30 π Μ β 1/β30 π Μ + 2/β30 π Μ Vector with magnitude 8 = 8 [5/β30 π Μ" β " 1/β30 π Μ" + " 2/β30 π Μ ] = ππ/βππ π Μ β π/βππ π Μ + ππ/βππ π Μ Hence, the required vector is 40/β30 π Μ β 8/β30 π Μ + 16/β30 π Μ

Ex 10.2

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About the Author

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.