In real life applications, we often deal with 3-dimensional objects and forces. These forces are described slightly differently than 2-dimensional forces. Let's take a look at how by looking at 2 cases.
In this case, we say that angles $ \theta $ and $ \phi $ are defined as below.
Force in 3D space where theta and phi are defined.
Using the notation from the image above, we can write our associated values for the force components.
[ 1 ]
$$ | \vec{F_z} | = | \vec{F} | \cos{\phi} $$
[ 2 ]
$$ | \vec{F_xy} | = | \vec{F} | \sin{\phi} $$
[ 3 ]
$$ | \vec{F_x} | = | \vec{F_xy} | \cos{\theta} = | \vec{F} | \sin{\phi}\cos{\theta} $$
[ 4 ]
$$ | \vec{F_y} | = | \vec{F_xy} | \sin{\theta} = | \vec{F} | \sin{\phi}\sin{\theta} $$
[ 5 ]
$$ | \vec{F} | = \sqrt{| \vec{F_x} |^2 + | \vec{F_y} |^2 + | \vec{F_z} |^2} $$
Now let's say we have three coordinate angles defined. These angles could be $ \alpha $, $ \beta $, and $ \gamma $ for example.
Force in 3D space where alpha, beta, and gamma are defined.
We can now use the image above to describe the force components:
[ 6 ]
$$ \vec{F_x} = F_x \hat{i} $$
[ 7 ]
$$ \vec{F_y} = F_y \hat{i} $$
[ 8 ]
$$ \vec{F_z} = F_z \hat{i} $$
[ 9 ]
$$ \vec{F} = F_x \hat{i} + F_y \hat{j} + F_z \hat{k} $$
These are a lot nicer looking than the previous case! We can also use a handy trick for if we are only given two of these angles:
[ 10 ]
$$ \cos^2{\alpha} + \cos^2{\beta} + \cos^2{\gamma} = 1 $$
Now let's say we have a unit vector, described as a directional vector divided by its magnitude, and we wanted to determine how a force would be distributed along this unit vector. We could define
[ 11 ]
$$ \hat{u}_{AB} = \frac{(x_2-x_1) \hat{i} + (y_2-y_1) \hat{j} + (z_2-z_1) \hat{k}}{\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}} $$
where $ \hat{u}_{AB} $ is our unit vector pointing from a point, A, in space with coordinates $ (x_1,y_1,z_1) $ to a point, B, in space with coordinates $ (x_2,y_2,z_2) $. Then we can say
[ 12 ]
$$ \vec{F} = F \hat{u}_{AB} $$ $$ \Rightarrow F(\frac{(x_2-x_1) \hat{i} + (y_2-y_1) \hat{j} + (z_2-z_1) \hat{k}}{\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}}) $$
A unit vector is never greater than 1, but it can range from 0 to 1. Think of a unit vector as a description of "how much" a vector is in a certain direction, whether that be the x, y, or z-direction. Above in Eq. (13), we are using this fact and multiplying the unit vector $ \hat{u}_{AB} $ by the magnitude of F to determine "how much" of the force is in each direction. This is an odd thing to visualize, but I have found drawing pictures always helps things click!