0
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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

$$|\vec{PA}|=\sqrt{\frac{\color{red}9}{4}+\frac{1}{4}+1}=\frac12\sqrt{14}$$

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.