Question

Find the shortest distance between the lines whose vector equtions are
$r=(1-t)\hat{i}+(t-2)\hat{j}+(2-2t)\hat{k}andr=(s+1)\hat{i}+(2s-1)\hat{j}-(2s+1)\hat{k}$

Answer 1

Equation of the given lines of the form
$r=(\hat{i}-2\hat{j}+3\hat{k})+t(-\hat{i}+\hat{j}-2\hat{k})andr=(\hat{i}-\hat{j}-\hat{k})+s(\hat{i}+2\hat{j}-2\hat{k})$
Compare these with $r=a_1+t{b}_{1}andr=a_2+s{b}_2$
$a_1=\hat{i}-2\hat{j}+3\hat{k},b_1=-\hat{i}+\hat{j}-2\hat{k}and{a}_2=\hat{i}-\hat{j}-\hat{k}{b}_2=\hat{i}+2\hat{j}-2\hat{k}$
Now, $a_2-a_1=(\hat{i}-\hat{j}-\hat{k})-(\hat{i}-2\hat{j}+3\hat{k})=\hat{j}-4\hat{k}$
and $b_1\times b_2=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -1& 1& -2\\ 1& 2& -2\end{array}\right|=(-2+4)\hat{i}-(2+2)\hat{j}+(-2-1)\hat{k}$
$=2\hat{i}-4\hat{j}-3\hat{k}$
$\Rightarrow |b_1\times b_2|=\sqrt{(2{)}^2+(-4{)}^2+(-3{)}^2}=\sqrt{4+16+9}=\sqrt{29}$
$\therefore$ Required shortest distance
$d=\left|\frac{(b_1\times b_2).(a_2-a_1)}{|b_1\times b_2|}\right|=\frac{|(2\hat{i}-4\hat{j}-3\hat{k}).(\hat{j}-4\hat{k})|}{\sqrt{29}}=\frac{|-4+12|}{\sqrt{29}}=\frac{|-4+12|}{\sqrt{29}}=\frac{8}{\sqrt{29}}$