Q15 (c) (ii)

$\mathbf c=\begin{bmatrix}1\\4\\0\end{bmatrix}+t\begin{bmatrix}-1\\1\\5\end{bmatrix}\text{ in parametric form is }\mathbf c=\begin{bmatrix}1-t\\4+t\\5t\end{bmatrix}$

If $T$ is any point on $c$, then the vector $\overrightarrow{OT}$ in terms of $t$ is
$\begin{bmatrix}1-t\\4+t\\5t\end{bmatrix}.$

By substituting the parametric equation into the result in part i)

$\overrightarrow{OS}=\frac{(1-t)+3(4+t)+(5t)}{11}\begin{bmatrix}1\\3\\1\end{bmatrix}$

$\overrightarrow{OS}=\frac{13+7t}{11}\begin{bmatrix}1\\3\\1\end{bmatrix}$

$\overrightarrow{TS}=\overrightarrow{OS}-\overrightarrow{OT}=\frac{13+7t}{11}\begin{bmatrix}1\\3\\1\end{bmatrix}-\begin{bmatrix}1-t\\4+t\\5t\end{bmatrix}$
$=\frac1{11}\begin{bmatrix}(13+7t)-11(1-t)\\3(13+7t)-11(4+t)\\(13+7t)-11(5t)\end{bmatrix}$
$=\frac1{11}\begin{bmatrix}2+18t\\-5+10t\\13-48t\end{bmatrix}$

1 – correct parametric representation of $T$ as a point on $c$
1 – correct calculation of $\overrightarrow{TS}$
1 – correct answer