To determine the unit vector in the y-z plane that is perpendicular to $\vec{A} = \hat{i} + \hat{j} + \hat{k}$, let the required unit vector be $\vec{u}$. Step 1: Define the vector $\vec{u}$ based on the given conditions. Since the vector is in the y-z plane, its x-component must be 0. Let $\vec{u} = y\hat{j} + z\hat{k}$. Step 2: Use the condition that $\vec{u}$ is a unit vector. The magnitude of a unit vector is 1. $$|\vec{u}| = \sqrt{y^2 + z^2} = 1$$ Squaring both sides gives: $$y^2 + z^2 = 1 \quad (*)$$ Step 3: Use the condition that $\vec{u}$ is perpendicular to $\vec{A}$. If two vectors are perpendicular, their dot product is 0. $$\vec{u} \cdot \vec{A} = 0$$ $$(0\hat{i} + y\hat{j} + z\hat{k}) \cdot (\hat{i} + \hat{j} + \hat{k}) = 0$$ $$(0)(1) + (y)(1) + (z)(1) = 0$$ $$y + z = 0$$ This implies: $$z = -y \quad (**)$$ Step 4: Solve the system of equations. Substitute equation $(*)$ into equation $()$: $$y^2 + (-y)^2 = 1$$ $$y^2 + y^2 = 1$$ $$2y^2 = 1$$ $$y^2 = \frac{1}{2}$$ $$y = \pm \sqrt{\frac{1}{2}} = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2}$$ Step 5: Find the corresponding values for $z$ and the unit vectors. 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 $\boxed{\frac{\sqrt{2}}{2}\hat{j} - \frac{\sqrt{2}}{2}\hat{k} \text{ and } -\frac{\sqrt{2}}{2}\hat{j} + \frac{\sqrt{2}}{2}\hat{k}}$.