Express position vector AB in cartesian vector form b Expres

Express position vector AB in cartesian vector form. b) Express position vector AC in cartesian vector form. c) Express the force F using cartesian vector notation. d) Determine the angle theta between cables AC and AB. Given: F = 12 lb a = 8 ft b = 10ft c = 8 ft d = 10 ft e = 4 ft f = 6 ft

Solution

Point A lies in x-y plane, B in y-z plane and C in x-z plane.
First, lets find the coordinates of points A, B and C.
A is in x-y plane => z_A = 0
X coordinate is distance from Y axis and vice-versa. We get A(f,c,0)
Similarly we can get B(0,e,d) and C(a,0,b)

Now vector AB = (ej + dk ) - (fi + cj) = -fi + (e-c)j + dk = -6i - 4j + 10k
Unit vector along AB = (-6i - 4j + 10k ) / |-6i - 4j + 10k | = (-3i - 2j + 5k)/(38)^1/2

Vector AC = (ai + bk) - (fi - cj) = 2i + 8j + 10k
Unit vector along AC = (2i + 8j + 10k) / |2i + 8j + 10k| = (i + 4j + 5k)/(42)^1/2

Force F is in direction AB and has magnitude of 12. F = 12(-3i - 2j + 5k)/(38)^1/2

Dot product : AB.AC = |AB||AC|cos(x) , where x is angle between AB and AC
hence cos(x) = (-6i - 4j + 10k).(2i + 8j + 10k)/(152)^1/2(168)^1/2 = 56/(152)^1/2(168)^1/2

cos(x) = 0.3504 => x = cos^-1(0.3504) = 69.48⁰