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Last updated at Dec. 24, 2019 by Teachoo

Transcript

Misc 6 Find a vector of magnitude 5 units, and parallel to the resultant of the vectors π β = 2π Μ + 3π Μ β π Μ and π β = π Μ β 2π Μ + π Μ. π β = 2π Μ + 3π Μ β π Μ π β = π Μ β 2π Μ + π Μ (π β + π β) = (2 + 1)π Μ + (3 β 2)π Μ + (β1 + 1)π Μ = 3π Μ + 1π Μ + 0π Μ Let π β = (π β + π β) β΄ π β = 3π Μ + 1π Μ + 0π Μ Magnitude of π β = β(32+12+02) |π β | = β(9+1) = β10 Unit vector in direction of π β = 1/|π β | Γ π β π Μ = 1/β10 Γ [3π Μ+1π Μ+0π Μ ] π Μ = π/βππ π Μ + π/βππ π Μ + 0π Μ Vector with magnitude 1 = 3/β10 π Μ + 1/β10 π Μ + 0π Μ Vector with magnitude 5 = 5 Γ [3/β10 " " π Μ" + " 1/β10 π Μ" + 0" π Μ ] = 15/β10 π Μ + 2/β10 π Μ + 0π Μ = 15/β10 π Μ + 2/β10 π Μ Rationalizing = 15/β10 Γ β10/β10 π Μ + 2/β10 "Γ " β10/β10 π Μ = (15β10)/10 π Μ + (2β10)/10 π Μ = (3β10)/2 π Μ + β10/5 π Μ Hence the required vector is (πβππ)/π π Μ + βππ/π π Μ

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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 9 years. He provides courses for Maths and Science at Teachoo.