5. In the tetrahedron $P-ABC$, edges $PA$, $AB$, and $AC$ are pairwise perpendicular, and $PA=AB=AC$. Points $E$ and $F$ are the midpoints of $AB$ and $PC$, respectively. Then the sine value of the angle $\theta$ formed by $EF$ and the plane $PBC$ is $\qquad$
5. $\frac{1}{3}$.

As shown in Figure 4, let $P A=$ $A B=A C=2$.
$$
\begin{array}{l}
\text { Then } E F=\sqrt{2+1}=\sqrt{3} . \\
\text { By } \frac{1}{2}(2 \sqrt{2})^{2} \frac{\sqrt{3}}{2} \times h \\
=\frac{1}{2} \times 2 \times 2 \times 2 \\
\Rightarrow h=\frac{4}{2 \sqrt{3}}=\frac{2}{\sqrt{3}} .
\end{array}
$$

Thus, $h_{E}=\frac{1}{\sqrt{3}}$.
Therefore, $\sin \theta=\frac{1}{\sqrt{3} \times \sqrt{3}}=\frac{1}{3}$.