Question

Find the shortest distance between the lines $\overrightarrow{r}=(4i-j)+\lambda (i+2j-3k)$ and $\overrightarrow{r}=(i-j+2k)+\mu (2i+4j-5k)$

• A. $\frac{2}{\sqrt{5}}$
• B. $\frac{2}{\sqrt{15}}$
• C. $\frac{6}{\sqrt{5}}$
• D. $\frac{6}{\sqrt{15}}$

Answer 1

The correct option is C $\frac{6}{\sqrt{5}}$

We know, the shortest distance between the lines
$r=a_1+\lambda b_1$ and $r=a_2+\lambda b_2$ is
$d=\left|\frac{(a_2-a_1)\cdot (b_1\times b_2)}{|b_1\times b_2|}\right|$

On comparing the given equations $\overrightarrow{r}=(4i-j)+\lambda (i+2j-3k)$ and $\overrightarrow{r}=(i-j+2k)+\mu (2i+4j-5k)$ with $r=a_1+\lambda b_1$ and $r=a_2+\lambda b_2$ respectively, we have

$a_1=4i-j,a_2=i-j+2k,b_1=i+2j-3k$ and $b_2=2i+4j-5k$
Now, $a_2-a_1=-3i+0j+2k$

and $b_1\times b_2=\left|\begin{array}{ccc}i& j& k\\ 1& 2& -3\\ 2& 4& -5\end{array}\right|=2i-j+0k$

$\therefore (a_2-a_1)\cdot (b_1\times b_2)=(-3i+0j+2k)\cdot (2i-j+0k)=-6$
and $|b_1\times b_2|=\sqrt{4+1+0}=\sqrt{5}$

$\therefore$ Shortest distance, $d=\left|\frac{(a_2-a_1)\cdot (b_1\times b_2)}{|b_1\times b_2|}\right|=\left|\frac{-6}{\sqrt{5}}\right|=\frac{6}{\sqrt{5}}$