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If we were to project the ecliptic out as an infinite plane, what visible star (besides the sun) would lie closest to it?

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Regulus ($\alpha$ Leo) is the one to beat, with ecliptic latitude $\beta$ = +0.46$^\circ$ and apparent V magnitude +1.4. Some dimmer stars are closer to the ecliptic, e.g. $\alpha$ Lib ($\beta$=+0.33$^\circ$, V=+2.7), $\delta$ Gem ($\beta$=-0.18$^\circ$, V=+3.5), and $\delta$ Cnc ($\beta$=+0.08$^\circ$, V=+3.9). For more suggestions, poke around in Stellarium with the Ecliptic turned on.

If you want linear distance from the ecliptic plane, multiply the distance from us by $\sin \beta$.

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HIP 76880 = $\kappa$ Librae (V=4.72) has an Ecliptic latitude of -0.019 degrees.

This is the winner amongst all stars in the Hipparco/Tycho catalogue with a Hipparcos magnitude <6.

If you mean physical distance, rather than angular distance, then the sine of the ecliptic latitude must be multiplied by a distance estimate for the star. This cannot be conclusive, the distances to many naked eye stars are very uncertain, a small fraction of Hipparcos parallaxes are too uncertain to be useful in this regard. However, from those with decent parallaxes then HIP 3765 = HD 4628 (V=5.72) has an absolute distance of 0.036 pc above the ecliptic plane.

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  • $\begingroup$ $\gamma^2$ Nor ecliptic latitude is -28.3 degrees. $\endgroup$
    – Mike G
    Commented May 7, 2016 at 20:42
  • $\begingroup$ @MikeG Aaaargh, I put in Galactic coords! $\endgroup$
    – ProfRob
    Commented May 7, 2016 at 22:17
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    $\begingroup$ Now fixed (I hope). $\endgroup$
    – ProfRob
    Commented May 7, 2016 at 23:06

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