Question

Find the shortest distance between the lines
$\left(a\right)\overrightarrow{r}=(8+3\lambda )\hat{i}-(9+16\lambda )\hat{j}+(10+7\lambda )\hat{k}$
and $\overrightarrow{r}=15\hat{i}+29\hat{j}+5\hat{k}+u\left(3\widehat{i+8\hat{j-5\hat{k}}}\right)$

Answer 1

$\begin{array}{c}\overrightarrow{r}=(8+3\lambda )\hat{i}-(9+16\lambda )\hat{j}+(10+7\lambda )\hat{k}\\ =\left(8\hat{i}-9\hat{j}+10\hat{k}\right)+\lambda \left(3\hat{i}-15\hat{j}+7\hat{k}\right)\\ \overrightarrow{r}=15\hat{i}+29\hat{j}+5\hat{k}+\mu (3\hat{i}+8\hat{j}-5\hat{k})\\ shortestdis\tan ce=\frac{\left|\left(-7\hat{i}-38\hat{j}+5\hat{k}\right).\left(\left(3\hat{i}-16\hat{j}+7\hat{k}\right)\times \left(3\hat{i}+8\hat{j}-5\hat{k}\right)\right)\right|}{\left|\left(3\hat{i}-16\hat{j}+7\hat{k}\right)\times \left(3\hat{i}+8\hat{j}-5\hat{k}\right)\right|}\\ d=\frac{\left|\left(-7\hat{i}-38\hat{j}+5\hat{k}\right).\left(24\hat{i}+36\hat{j}+72\hat{k}\right)\right|}{\sqrt{{\left(24\right)}^2+{\left(36\right)}^2+{\left(72\right)}^2}}\\ =\frac{\left|-168+\left(-1368\right)+360\right|}{\sqrt{576+1296+5184}}\\ d=\frac{1176}{84}\\ =14\end{array}$