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Let the applied force be $\mathbf{f}=f\hat{\mathbf{z}}$. A segment with polar angle $\theta$ has energy $-fb\cos \theta$. Independent 3D orientations give the one-link partition functionTaking the force derivative of $k_{B}T\ln Z_N$ yields the mean end-to-end projection,For $u\ll 1$, $\coth u-1/u=u/3+O(u^3)$, soIsotropy supplies the vector relation. It is the mean vector or its projection, not the mean magnitude of a fluctuating chain; at zero force $⟨\mathbf{R}⟩=0$ while $⟨R^2⟩=N{b}^2$ remains positive. The last panel of the model figure in solution shows the exact orientational response and its weak-force tangent.