To find a vector that has the same magnitude as vector $$\vec{A}$$ and is parallel to vector $$\vec{B}$$, we use the formula:

$$\vec{N} = |\vec{A}| \hat{B}$$

First, let's find the magnitude of $$\vec{A}$$, which is $$|\vec{A}|$$.

$$|\vec{A}| = \sqrt{3^2 + 4^2}=5$$

We're given that $$\vec{B} = 4 \hat{i}+3 \hat{j}$$, so to make $$\vec{N}$$ parallel to $$\vec{B}$$ and have it have the same magnitude as $$\vec{A}$$, they should be the same when $$\vec{N}$$'s magnitude is divided by 5, since $$|\vec{A}|=5$$.

So, $$\vec{N} = \frac{5(4\hat{i}+3\hat{j})}{5} = 4\hat{i}+3\hat{j}$$.

This means the $$x$$ component of the vector in the first quadrant is 4, and the $$y$$ component is given as 3. So, $$x=4$$.