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How do you calculate the components, and the magnitude of projection in a 3D force vector system?

A 3D force F=100kN is applied at point A (1, 2, 2) and directed towards point B (3, 5, 4). Point C has x, y and z coordinates of (-1, 6, -1).

1) What are the x, y and z components of the force F?

2) What is the magnitude of the projection of the force onto the direction of BC?

More interested in the method rather than the answer, thanks!

1 Answer

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  • Nick
    Lv 6
    7 years ago
    Favorite Answer

    1) Work out the unit vector between the points A and B:

    AB^ = AB/|AB| = ( (3,5,4) - (1,2,2) )/|AB| = (2,3,2)/|AB|

    |AB| is the length of the vector AB and is given by Pythagoras:

    |AB| = sqrt( 2^2 + 3^2 + 2^2 ) = sqrt(17)

    AB^ = (2,3,2)/sqrt(17)

    The force has a magnitude of |F| = 100x10^3 N and has a vector which is:

    F = |F| AB^ = 100x10^3 (2,3,2)/sqrt(17)

    = (10^5/sqrt(17))(2,3,2)

    = ( 2x10^5/sqrt(17) , 3x10^5/sqrt(17) , 2x10^5/sqrt(17) )

    You can work out the square-root here or just leave it in exact form.

    2) The direction BC is given by the unit vector:

    BC^ = BC/|BC| = ( (-1,6,-1) - (3,5,4) )/|BC| = (-4,1,-5)/|BC|

    The magnitude |BC| given by:

    |BC| = sqrt(4^2 + 1^2 + 5^2) = sqrt(42)

    BC^ = (-4,1,-5)/sqrt(42)

    To find the component of F in the direction BC^ we take the dot product:

    F.BC^ = (10^5/sqrt(17*42)) (2,3,2).(-4,1,-5)

    = 10^5/sqrt(714) ((2*-4) + (3*1) + (2*-5))

    = -15x10^5/sqrt(714)

    = 5.6x10^4 N

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