Homework and Quiz Solutions

Homework and Quiz Solutions - Biomechanics, 3150...

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Biomechanics, 3150 Instructor, Shiva Kotha skotha@engr.uconn.edu Homework 1 Question 1) Use the figure below to answer the following questions. 1.1) What are the unit vectors along the X, Y, Z axis? Unit vectors along: X axis – i or (1, 0, 0) or i + 0j + 0k Y axis – j or (0, 1, 0) or 0i + j + 0k Z axis – k or (0, 0, 1) or 0i + 0j + k 1.2) What are the unit vectors along OA, OB, DE, BA, and CB? Vectors along: Unit vectors along: OA = A – O = (10-0) i + (2-0) j + (8-0) k OA = (10i + 2j + 8k)/ [sqrt(10 2 + 2 2 + 8 2 )] OA = 10i + 2j + 8k OA = 0.77i + 0.15j + 0.62k OB = 3i + 2j + 10k OB = 0.28i + 0.19j + 0.94k DE = E – D = (-2-2)i + (6-12)j + (5-0)k = -4i - 6j +5k DE = -0.46i - 0.68j + 0.58k BA = 7i -2k BA = 0.96i – 0.27k CB = -i - 4j + 9k CB = -0.1i - 0.4j + 0.91k 1.3) What is the angle between vector CD and vector CB? Angle between vectors, ‘θ’, is given by the dot product. CD.CB = magnitude(CD) magnitude(CB) cos(θ) Vector CD = -2i + 6j –k magnitude(CD) = sqrt[(-2) 2 + 6 2 + (-1) 2 ] = 6.4 Vector CB = -i - 4j + 9k magnitude(CB) = sqrt[(-1) 2 + (-4) 2 + 9 2 ] = 9.9 Dot product of CD and CB CD.CB = (-2)*(-1) + (6)*(-4) + (-1)*(9) = 2 - 24 - 9 = -31 cos (θ) = -31/(6.4 * 9.9), implies θ = cos -1 (-0.49), or θ = 119.3° Note: Figure is not to scale
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Biomechanics, 3150 Instructor, Shiva Kotha skotha@engr.uconn.edu Homework 1 1.4) Find the direction cosine of vector OE with the X, Y, and Z axis (i.e. angle that vector OE makes with the X, Y, and Z axis). Verify your answer by ensuring that the square of the cosines of the angles is 1. Vector OE = -2i + 6j + 5k Magnitude of Vector OE = sqrt[ (-2) 2 + 6 2 + 5 2 ] = 8.1 Assume that α, β, and γ are angle that vector OE makes with the X, Y, and Z axis respectively. cos α = -2/8.1; cos β = 6/8.1; and cos γ = 5/8.1 OR α = 104.2; β = 42.2; γ = 51.9 Verify using: cos 2 α + cos 2 β + cos 2 γ = 1; 1.5) Find the projection of vector CD on vector CE. Vector CD = -2i + 6j –k Vector CE = -6i + 0j +4k Magnitude of vector CE = sqrt[(-6) 2 + 0 2 + 4 2 ] = sqrt(52)=7.2 Unit vector along CE = M = [-6i + 0j + 4k]/sqrt(52) = -0.83i + 0j + 0.55k Projection of vector CD on vector CE is given by the dot product. [CD.M]
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Homework and Quiz Solutions - Biomechanics, 3150...

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