Question

Find the shortest distance between the lines whose vector equations are $r=(\hat{i}+2\hat{j}+3\hat{k})+\lambda (\hat{i}-3\hat{j}+2\hat{k})$
and $r=(4\hat{i}+5\hat{j}+6\hat{k})+\mu (2\hat{i}+3\hat{j}+\hat{k})$

Answer 1

The given lines are $r=\hat{i}+2\hat{j}+3\hat{k}+\lambda (\hat{i}-3\hat{j}+2\hat{k})$
and $r=4\hat{i}+5\hat{j}+6\hat{k}+\mu (2\hat{i}+3\hat{j}+\hat{k})$
Comparing the given equations with $r=a_1+\lambda b_{1}andr=a_2+\mu b_2$, we obtain
$a_1=\hat{i}+2\hat{j}+3\hat{k},b_1=\hat{i}-3\hat{j}+2\hat{k}and{a}_2=4\hat{i}+5\hat{j}+6\hat{k},b_2=2\hat{i}+3\hat{j}+\hat{k}Here,a_2-a_1=(4\hat{i}+5\hat{j}+6\hat{k})-(\hat{i}+2\hat{j}+3\hat{k})=3\hat{i}+3\hat{j}+3\hat{k}$
Also, $b_1\times b_2=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -3& 2\\ 2& 3& 1\end{array}\right|=\hat{i}(-3-6)-\hat{j}(1-4)+\hat{k}(3+6)=-9\hat{i}+3\hat{j}+9\widehat{[}k]$
$\Rightarrow |b_1\times b_2|=\sqrt{(-9{)}^2+(3{)}^2+(9{)}^2}=\sqrt{81+9+81}=\sqrt{171}=3\sqrt{19}$
$\therefore$ Required shortest distance
$d=\left|\frac{(b_1\times b_2).(a_2-a_1)}{|b_1\times b_2|}\right|=\frac{|(-9\hat{i}+3\hat{j}+9\hat{k})3(3\hat{i}+3\hat{j}+3\hat{k})|}{3\sqrt{19}}=\frac{|-9\times 3+3\times 3+9\times 3|}{3\sqrt{19}}=\frac{-27+9+27}{3\sqrt{19}}=\frac{9}{3\sqrt{19}}=\frac{3}{\sqrt{19}}$
Therefore, the shortest distance between the two given lines is $\frac{3}{\sqrt{19}}$ units