Answer
$\frac{\vec{A}}{A}$, where A is the magnitude of $\vec{A}$, will always be a unit vector because it will always have a magnitude of one and a direction in the original direction of $\vec{A}$
$\frac{\vec{A}}{A} \cdot \hat{i} = |1||1|cos(\theta) = cos(\theta)$
Work Step by Step
$\frac{\vec{A}}{A} = \frac{A_1 \hat{i} + A_2 \hat{j} + A_3 \hat{k}}{\sqrt{A_1^2 + A_2^2 + A_3^2}}$ has magnitude $\frac{(A_1^2 + A_2^2 + A_3^2)}{A_1^2 + A_2^2 + A_3^2} = 1$
The direction is still that of $\vec{A}$ because the magnitude is a scalar value; dividing by a scalar changes the length of the vector but does not affect the direction.
By the definition of the dot product:
$\frac{\vec{A}}{A} \cdot \hat{i} = |\frac{\vec{A}}{A}||\hat{i}|cos(\theta) = |1||1|cos(\theta) = cos(\theta)$
Here $\theta$ is the angle between our unit vector $\hat{A}$ and the x-axis. It is called the direction cosine for that axis because it denotes the angle between the +x axis and $\hat{A}$
