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Determine the angle 8 between cables AB and AC. Suppose that a-1.9 m and b-1.9 m. (Figure Express your answer using three sig

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Answer #1

Coordinate of A=(3,0,0)

Coordinate of B=(0,-0.75,1.9)

Coordinate of C=(0,1.9,1.5)

Vector AB=(0-3)i+(-0.75-0)j+(1.9-0)k=-3i-0.75j+1.9k

Vector AC=(0-3)i+(1.9-0)j+(1.5-0)k=-3i+1.9j+1.5k

Angle between two vectors is given by

arccos((AB.AC)/magnitude(AB) magnitude (AC))

AB.AC=(-3)(-3)+(-0.75)(1.9)+(1.9)(1.5)=10.425

Magnitude of AB=sqrt(3²+0.75²+1.9²)=3.63

Magnitude of AC=sqrt(3²+1.9²+1.5²)=3.85

Angle=arccos(10.425/(3.63*3.85))=41.8°

This is the required answer

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