= Solution
Divide the velocity by the <speed> calculated above. The <unit tangent vector> is
$$
\boxed{T(t)=\frac1{\sqrt2}\left(\frac{1-t^2}{1+t^2},\frac{2t}{1+t^2},1\right).}
$$
Its derivative is
$$
T'(t)=\frac1{\sqrt2(1+t^2)^2}\bigl(-4t,2(1-t^2),0\bigr),
\qquad |T'(t)|=\frac{\sqrt2}{1+t^2}>0.
$$
Because $ds/dt=\sqrt2(1+t^2)>0$, the <principal normal vector> has the direction of $T'(t)$, giving
$$
\boxed{N(t)=\left(-\frac{2t}{1+t^2},\frac{1-t^2}{1+t^2},0\right).}
$$
This is a <unit vector>, and $N\cdot e_z=0$ for every real $t$. The <curvature of a space curve> is $|T'|/(ds/dt)=1/(1+t^2)^2>0$, so the <principal normal vector> is defined everywhere on this <curve>. The same perpendicularity follows from <constant tangent component forces perpendicular principal normals>: $T\cdot e_z=1/\sqrt2$ is constant, and differentiating this identity forces the normal component along $e_z$ to vanish.
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