Distance between line and a point Consider the points (1,2,-1) and (2,0,3).
(a) Find a vector equation of the line through these points in parametric form.
(b) Find the distance between this line and the point (1,0,1). (Hint: Use the parametric form of the equation and the dot product)
I have solved (a), Forming:
Vector equation: (1,2,-1)+t(1,-2,4)
x=1+t
y=2-2t
z=-1+4t
However, I'm a little stumped on how to solve (b).

Using this formula and your computations from (a), we get the expression for distance $d$:
$$
d = 
\frac{\left\| \ 
\
\left(\ 
\begin{bmatrix}
1 \\ 2 \\ -1
\end{bmatrix}
- 
\begin{bmatrix}
1 \\ 0 \\ 1
\end{bmatrix}
\ 
\right)
\times
 \, 
\begin{bmatrix}
1 \\ -2 \\ 4
\end{bmatrix}
 \ \ 
\right\|}
{
 \begin{Vmatrix}
1 \\ -2 \\ 4
\end{Vmatrix}
}
=
\frac{\left\| \ 
\
\begin{bmatrix}
0 \\ 2 \\ -2
\end{bmatrix} 
\times
\begin{bmatrix}
1 \\ -2 \\ 4
\end{bmatrix}
 \ \ 
\right\|}
{
 \sqrt{1 + 4 + 16}
}
=
$$
$$
=
\frac{1}{\sqrt{21}}
\begin{Vmatrix}
 \vec{\mathbf{i}} &  \vec{\mathbf{j}} &  \vec{\mathbf{k}} \\
0 & 2 & -2 \\
1 & -2 & 4
\end{Vmatrix}
=
\frac{1}{\sqrt{21}}
\begin{Vmatrix}
2 \cdot 4 - (-2) \cdot (-2) \\
(-2)\cdot 1 - 0 \cdot 4 \\
0 \cdot (-2 ) - 1 \cdot 2
\end{Vmatrix}
=
\frac{1}{\sqrt{21}}
\begin{Vmatrix}
4 \\
-2 \\
-2
\end{Vmatrix}
=
$$
$$
= 
\frac{\sqrt{16+4+4}}{\sqrt{21}} = \sqrt{\frac{24}{21}}= \sqrt{\frac{8}{7}} = \frac{2\sqrt{2}}{\sqrt{7}} \approx 1.069
$$
