= Solution
Using the needle definitions in the question-level solution, $\mathbf u\cdot\mathbf e=u_z=\cos\theta$. The scattered field is parallel to $\mathbf v=u_x\hat{\mathbf x}+u_z\hat{\mathbf z}$, and
$$
\boxed{\frac{I_a}{KI_0}=u_z^2(1-u_y^2)=\cos^2\theta\left(1-\sin^2\theta\sin^2\phi\right).}
$$
The light is \b[linearly polarized] along the needle's projection into the $xz$ observation plane. Its <Jones vector> in the $(x,z)$ basis is $(u_x,u_z)^{\mathsf T}/\sqrt{1-u_y^2}$ whenever the intensity is nonzero. A needle perpendicular to the incident <electric field>, or pointing directly at the observer, has zero leading-order signal.
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