Answer
(a) its curvature is zero
(b) the space curve is a line
(c) the space curve lies in a plane
Work Step by Step
(a) A space curve which has a constant unit tangent vector ${\bf{T}}$ has unit speed. By definition, it is parametrized by arc length. By definition of curvature $\kappa \left( s \right) = ||\frac{{d{\bf{T}}}}{{ds}}||$, its curvature is zero.
(b) The normal vector at $t$ is given by ${\bf{N}}\left( t \right) = \frac{{{\bf{T}}'\left( t \right)}}{{||{\bf{T}}'\left( t \right)||}}$. If the normal vector is constant, then ${\bf{T}}'\left( t \right)$ is constant. Thus, the unit tangent vector is a linear function of $t$. Hence, the space curve is a line.
(c) The binomial vector ${\bf{B}}$ is perpendicular to the osculating plane formed by ${\bf{T}}$ and ${\bf{N}}$. If ${\bf{B}}$ is constant, then the space curve lies in a plane.
