# Finding the distance between two parallel 3D vectors

Find the distance between the pairs of parallel lines: $$r_a=\frac{x-2}{1}=\frac{y-1}{-1}=\frac{z-3}{2}\\ r_b=\frac{x+1}{1}=\frac{y-3}{-1}=\frac{z-1}{2}$$

Let $A(2,1,3)$ sit on $r_a$ and let $P$ be a point where the perpendicular of $A$ meets $r_b$. Let $P$ have the position vector $p$:

$$p = \begin{pmatrix}-1 \\ 3\\1\end{pmatrix} + \lambda\begin{pmatrix}1 \\ -1\\2\end{pmatrix}=\begin{pmatrix}-1+\lambda \\ 3-\lambda\\1+2\lambda\end{pmatrix} \\ \vec{PA}=a-p=\begin{pmatrix}3-\lambda \\ -2+\lambda\\2-2\lambda\end{pmatrix}$$

But $\vec{PA}$ is perpendicular to $r_b$ and $r_b$ is parallel to $(1,-1,2)$, so:

$$\begin{pmatrix}3-\lambda \\ -2+\lambda\\2-2\lambda\end{pmatrix}\cdot\begin{pmatrix}1\\-1\\2\end{pmatrix}=0\\ 3-\lambda+2-\lambda+4-4\lambda = 0\\ \lambda = \frac{3}{2}$$ Therefore: $$\vec{PA}=\begin{pmatrix}\frac{1}{2} \\ -\frac{1}{2}\\-1\end{pmatrix} \\ |\vec{PA}|=\sqrt{\frac{1}{4}+\frac{1}{4}+1}=\frac{\sqrt3}{\sqrt2}$$ But the answer given is $\frac{1}{2}\sqrt{14}$. Where am I going wrong?

$$\vec{PA}=\begin{pmatrix}3-\lambda \\ -2+\lambda\\2-2\lambda\end{pmatrix}$$
Substituting $\lambda=\dfrac{3}{2}$ gives: $$\vec{PA}=\begin{pmatrix}3-1\frac12\\-2+1\frac12\\2-2\cdot1\frac12\end{pmatrix}=\begin{pmatrix}\color{red}1\frac{1}{2} \\ -\frac{1}{2}\\-1\end{pmatrix}$$
$$|\vec{PA}|=\sqrt{\frac{\color{red}9}{4}+\frac{1}{4}+1}=\frac12\sqrt{14}$$