Let the vector be $\vec{r}$. Let $\alpha$, $\beta$, and $\gamma$ be the angles made by the vector $\vec{r}$ with the positive x-axis, y-axis, and z-axis, respectively.

Given, $\alpha = \frac{\pi}{4}$ and $\beta = \frac{\pi}{4}$. We need to find $\gamma$.

We know that the sum of the squares of the direction cosines is equal to 1, i.e.,

$\cos^2{\alpha} + \cos^2{\beta} + \cos^2{\gamma} = 1$

Substituting the given values, we have:

$\cos^2{\frac{\pi}{4}} + \cos^2{\frac{\pi}{4}} + \cos^2{\gamma} = 1$

$(\frac{1}{\sqrt{2}})^2 + (\frac{1}{\sqrt{2}})^2 + \cos^2{\gamma} = 1$

$\frac{1}{2} + \frac{1}{2} + \cos^2{\gamma} = 1$

$1 + \cos^2{\gamma} = 1$

$\cos^2{\gamma} = 0$

$\cos{\gamma} = 0$

Therefore, $\gamma = \frac{\pi}{2}$