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?


It's just a simple arithmetic error:

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

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


  • $\begingroup$ D'oh :) thanks! $\endgroup$
    – hohner
    Oct 27 '14 at 18:54

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