= Solution
Take positive $x$ horizontally toward the downhill end of the plane, with $0\leq\theta<\pi/2$. The <normal force> is $N=mg\cos\theta$, since there is no <acceleration> perpendicular to the stationary plane. Along the plane, the only tangential <force> is $mg\sin\theta$, so <Newton's second law> gives $a_\parallel=g\sin\theta$.
Its horizontal projection is therefore
$$
\boxed{a_x=g\sin\theta\cos\theta.}
$$
Equivalently, the <normal force> has horizontal component $N\sin\theta$ while gravity has none, giving $ma_x=N\sin\theta$. For a plane whose height increases in the positive $x$ direction, the answer instead carries a minus sign. A horizontal plane gives zero <acceleration>; the horizontal projection also tends to zero for a vertical plane.
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