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).
 A: You can use a formula, although I think it's not too difficult to just go through the steps.  I would draw a picture first:

You are given that $\vec{p} = (1,0,1)$ and you already found $\vec{m} = (1, -2, 4)$ and $\vec{l}_0 = (1,2,-1)$.  Now it's a matter of writing an expression for $\vec{l}(t) - \vec{p}_0$:
\begin{align}
\vec{l}(t) - \vec{p}_0 =&\ (\ (t + 1) - 1\ ,\ (-2t + 2) - 0\ ,\ (4t - 1) - 1\ )\\
=&\ (\ t\ ,\ -2t + 2\ ,\ 4t - 2\ )
\end{align}
Now you dot this with the original slope of the line (recall that $\vec{l}(t) - \vec{p}_0$ is the slope of the line segment connecting the point and the line).  When this dot product equals zero, you have found $t_0$ and thus $\vec{x}_0$:
\begin{align}
\vec{m} \circ (\vec{l}(t) - \vec{p}_0) =&\ (1,-2,4)\circ(\ t\ ,\ -2t + 2\ ,\ 4t - 2\ ) \\
=&\ t + 4t - 4 + 16t - 8 \\
=&\ 21t - 12
\end{align}
Setting this to $0$ gives that $21t_0 - 12 = 0 \rightarrow t_0 = \frac{4}{7}$.  This gives the point $\vec{x}_0$ as:
\begin{align}
\vec{x}_0 =&\ \vec{l}(t_0) =  (\ \frac{4}{7} + 1\ ,\ -\frac{8}{7} + 2\ ,\ \frac{16}{7} - 1\ ) \\
=&\ \frac{1}{7}(11, 6, 9)
\end{align}
So finally the distance would be the distance from $\vec{p}_0$ to $\vec{x}_0$:
\begin{align}
d =&\ \sqrt{\left(\frac{11}{7} - 1\right)^2 + \left(\frac{6}{7} - 0\right)^2 + \left(\frac{9}{7} - 1\right)^2}\\
=&\ \sqrt{\left(\frac{4}{7}\right)^2 + \left(\frac{6}{7}\right)^2 + \left(\frac{2}{7}\right)^2} \\
=&\ \frac{1}{7}\sqrt{4^2 + 6^2 + 2^2}\\
=&\ \frac{1}{7}\sqrt{56} \\
=&\ \frac{2}{7}\sqrt{14} 
\end{align}
...or perhaps $\sqrt{\frac{8}{7}}$ is more appealing.
Extra Info
There's no need to worry about whether or not my 2D picture is really representative--it is.  No matter how high the dimensions of the problem, the problem itself can always be mapped to exactly 2 dimensions unless the point is on the line--then it's a 1 dimensional problem--which of course we can represent in 2 dimensions just as we can represent this 2 dimensional problem in much higher ones.
A: A line through the points $p_1$ and $p_2$ can be written as
$$
\bbox[5px,border:2px solid #00A000]{p=p_1+(p_2-p_1)t}\tag{1}
$$
The distance from the line in $(1)$ is given by
$$
\bbox[5px,border:2px solid #C0A000]{\left|\,(p-p_1)-\frac{(p-p_1)\cdot(p_2-p_1)}{|p_2-p_1|^2}(p_2-p_1)\,\right|}\tag{2}
$$

Plugging in the values for the points $p_1=(1,2,-1)$, $p_2=(2,0,3)$, and $p=(1,0,1)$ into $(1)$ and $(2)$, we get that the line sought is
$$
\bbox[5px,border:2px solid #00A000]{(1,2,-1)+(1,-2,4)t}\tag{3}
$$
and the distance sought is
$$
\begin{align}
\left|\,(0,-2,2)-\frac{(0,-2,2)\cdot(1,-2,4)}{|(1,-2,4)|^2}(1,-2,4)\,\right|
&=\left|\,(0,-2,2)-\frac{12}{21}(1,-2,4)\,\right|\\[6pt]
&=\left|\,\frac17(-4,-6,-2)\,\right|\\
&=\bbox[5px,border:2px solid #C0A000]{\frac{2\sqrt{14}}7}\tag{4}
\end{align}
$$

Justification of $\boldsymbol{(2)}$
Note that
$$
(p-p_1)-\frac{(p-p_1)\cdot(p_2-p_1)}{|p_2-p_1|^2}(p_2-p_1)\tag{5}
$$
and
$$
\frac{(p-p_1)\cdot(p_2-p_1)}{|p_2-p_1|^2}(p_2-p_1)\tag{6}
$$
are perpendicular (their dot product is $0$) and sum to $p-p_1$. Thus, they form the triangle

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