Define the vector u based on the given conditions.

Since the vector is in the y-z plane, its x-component must be 0.
Let $\vec{u} = y\hat{j} + z\hat{k}$.

The magnitude of a unit vector is 1.

Squaring both sides gives:

If two vectors are perpendicular, their dot product is 0.

This implies:

Substitute equation $(**)$ into equation $(*)$:

Case 1: If $y = \frac{\sqrt{2}}{2}$
Using $z = -y$, we get $z = -\frac{\sqrt{2}}{2}$.
So, the unit vector is $\vec{u}_1 = \frac{\sqrt{2}}{2}\hat{j} - \frac{\sqrt{2}}{2}\hat{k}$.

Case 2: If $y = -\frac{\sqrt{2}}{2}$
Using $z = -y$, we get $z = -(-\frac{\sqrt{2}}{2}) = \frac{\sqrt{2}}{2}$.
So, the unit vector is $\vec{u}_2 = -\frac{\sqrt{2}}{2}\hat{j} + \frac{\sqrt{2}}{2}\hat{k}$.

Both vectors satisfy the given conditions.

The unit vectors are $\boxed{\frac{\sqrt{2}}{2}\hat{j} - \frac{\sqrt{2}}{2}\hat{k} \text{ and } -\frac{\sqrt{2}}{2}\hat{j} + \frac{\sqrt{2}}{2}\hat{k}}$.