Answer
False
Work Step by Step
$$\eqalign{
& {\text{Let }}{\bf{r}}\left( t \right) = \left\langle {3\cos t,2\sin t,t - 2} \right\rangle {\text{ and }}{\bf{u}}\left( t \right) = \left\langle {4\sin t, - 6\cos t,{t^2}} \right\rangle \cr
& {\text{Find }}{\bf{r}}\left( t \right) \cdot {\bf{u}}\left( t \right) \cr
& {\bf{r}}\left( t \right) \cdot {\bf{u}}\left( t \right) = \left\langle {3\cos t,2\sin t,t - 2} \right\rangle \cdot \left\langle {4\sin t, - 6\cos t,{t^2}} \right\rangle \cr
& {\bf{r}}\left( t \right) \cdot {\bf{u}}\left( t \right) = \left( {3\cos t} \right)\left( {4\sin t} \right) + \left( {2\sin t} \right)\left( { - 6\cos t} \right) + \left( {t - 2} \right){t^2} \cr
& {\bf{r}}\left( t \right) \cdot {\bf{u}}\left( t \right) = 12\cos t\sin t - 12\sin t\cos t + {t^3} - 2{t^2} \cr
& {\bf{r}}\left( t \right) \cdot {\bf{u}}\left( t \right) = {t^3} - 2{t^2} \cr
& {\text{The statement is false, the dot product is a scalar - valued function}} \cr} $$
