Determine the coordinate direction angles of force F_1

#### Solution:

### We will start off by drawing a vector diagram to help us visualize the steps better like so:

##### Note that we only focus on force F_1 since the question only asks us to determine the coordinate direction angles of force F_1. Thus, we only drew the vector components of force F1. No need to draw unnecessary components since that will only make it more difficult for us to visualize the steps.

### Let us now look at the components of force F_1.

### We can write each component like so:

### (F_1)_x=600(\frac{4}{5})\,\text{cos}\,30^0 N

### (F_1)_y=600(\frac{4}{5})\,\text{sin}\,30^0 N

### (F_1)_x=600(\frac{3}{5}) N

### We can now write force F_1 in Cartesian vector form like so:

### F_1=600\left\{\frac{4}{5}\,\text{cos}\,30^0(+i)\,+\,\frac{4}{5}\,\text{sin}\,30^0(-j)\,+\,\frac{3}{5}(+k)\right\}

### F_1=600\left[0.6928{i}-0.4{j}+0.6{k}\right] N

##### (the signs for i,j, and k can be seen looking at the diagram. We can see that the j component is in the negative direction, thus, we can write it as -j)

### Now, we need to find the unit vector. To do so, remember that we simply divide the Cartesian vector form by the value of F_1 which in this case is 600 N.

### (u_F)_1=\frac{600(0.6928{i}-0.4{j}+0.6{k})\,\text{N}}{600\,\text{N}}

### (u_F)_1=0.6928{i}-0.4{j}+0.6{k}

### The last step is to find the coordinate direction angles.

### \alpha=\text{cos}^{-1}((u_{F})_{1})_x

### \alpha=\text{cos}^{-1}(0.6928) =46.1^0

### \beta=\text{cos}^{-1}((u_{F})_{1})_y

### \beta=\text{cos}^{-1}(-0.4) =114^0

### \gamma=\text{cos}^{-1}((u_{F})_{1})_z

### \gamma=\text{cos}^{-1}(0.6) =53.1^0

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