Given vector $\vec{a}=(1, \sqrt{3})$ and the angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{6}$, find the projection of $\vec{a}$ on $\vec{b}$.

1. Problem Description

Given vector $\vec{a}=(1, \sqrt{3})$ and the angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{6}$, find the projection of $\vec{a}$ on $\vec{b}$.

2. Solution Steps

Let the projection of $\vec{a}$ on $\vec{b}$ be $proj_{\vec{b}} \vec{a}$.

The formula for the projection of $\vec{a}$ on $\vec{b}$ is:

$proj_{\vec{b}} \vec{a} = |\vec{a}| \cos{\theta}$, where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$.

We are given that $\vec{a} = (1, \sqrt{3})$. Therefore, the magnitude of $\vec{a}$ is:

$|\vec{a}| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2$.

The angle between $\vec{a}$ and $\vec{b}$ is given as $\theta = \frac{\pi}{6}$.

Therefore, $\cos{\frac{\pi}{6}} = \frac{\sqrt{3}}{2}$.

Now, we can compute the projection of $\vec{a}$ on $\vec{b}$:

$proj_{\vec{b}} \vec{a} = |\vec{a}| \cos{\theta} = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3}$.

3. Final Answer

The projection of $\vec{a}$ on $\vec{b}$ is $\sqrt{3}$.