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Homework Help: Three Dimensional Vectors

  1. Oct 8, 2013 #1
    I need to determine the magnitude of resultant force R
    And, determine the coordinate direction angles α,β, γ of force R

    I just can't seem to understand 3-d vectors. I'm fine with 2-d, but this is so different to me. The answers are given in bold I just don't know how to get there. I'd really appreciate any help/explanations. I've asked multiple people for help and they were all confused.
    I would ask my professor, but he has a very strong accent and is difficult to understand.

    Thanks so much!
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution

    Attached Files:

  2. jcsd
  3. Oct 8, 2013 #2
    In your attachment, the answer is (38.22i + 67.99j + 17.5k). In three dimensional vectors, i, j, and k are simply used to describe the axes. i=x , j=y, k=z . So you have a magnitude of 38.22 along the x-axis, 67.99 along the y-axis, and 17.5 along the z-axis.

    It's very similar to 2-d vectors. For example, in 2 dimensional kinematics, a ball shot into the air can easily be broken down into a velocity along the y-axis, and a velocity in the x-axis. Visualizing this, you'll have a ball going up and down, as well as left and right; much like being stuck in a piece of paper, you can go up, down, left, and right.

    Adding the z component, you can imagine the ball also being allowed to go in and out of the page. In the diagram, you can see the axes that are labeled x, y, and z. The x and y parts are the same exact axes you would see when you plot something like y=2x+3 . This will essentially be the flat piece of paper that you would work on in two dimensional kinematics with the ball moving around. The z component will be the aspect of coming in and out of that page. (the graph is just conventionally rotated so you can see all three components, x, y, and z all at once. )


    There is an imagine to try and help you visually follow. As you can separate the components in the x and y directions in 2-d kinematics, you can do the same here.

    the magnitude in the x-direction is how much of the force is along the horizontal (or x-axis)

    the magnitude in the y-direction is how much the force is along the vertical (or y-axis)

    the magnitude in the z-direction is how much the force is coming into, or out of the page, (or the z-axis)

    I apologize if this confuses you further. You seemed to have a good grasp on the 2-d vectors, and wanted to build off of that foundation. If you have any questions or would like more clarification, please feel free to message me. I'll be glad to help.
    Last edited: Oct 8, 2013
  4. Oct 8, 2013 #3
    To find the resulting Force, break down each force into its separate components. the force along x, the force along y, the force along z. now simply add these together.

    Fx = Fx,1 + Fx,2

    Fy = Fy,1 + Fy,2

    Fz = Fz,1 + Fz,2

    From these, you should get the answer. (38.22x + 67.99y + 17.5z)

    Fx = 38.22
    Fy = 67.99
    Fz = 17.5

    To find the magnitude of the combined Force, in 2-dimensions, you just need to find the hypotenuse.

    l F l = √(fx)2 + (fy)2

    to find the magnitude in 3-dimensions, it's easily the same concept

    l F l = √(fx)2 + (fy)2 + (fz)2
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