Question

Find the shortest distance between the given lines by vector method.
$\overrightarrow{r}=3\hat{i}+8\hat{j}+3\hat{k}+\lambda (3\hat{i}-\hat{j}+\hat{k})$ and $\overrightarrow{r}=-3\hat{i}-7\hat{j}+6\hat{k}+\mu (-3\hat{i}+2\hat{j}+4\hat{k})$.

Answer 1

Given that lines:
$\overrightarrow{r}=3\hat{i}+8\hat{j}+3\hat{k}+\lambda (3\hat{i}-\hat{j}+\hat{k})$
and $\overrightarrow{r}=-3\hat{i}-7\hat{j}+6\hat{k}+\mu (-3\hat{i}+2\hat{j}+4\hat{k})$.
$\therefore a_1=3\hat{i}+8\hat{j}+3\hat{k},b_1=3\hat{i}-\hat{j}+\hat{k}$
$a_2=-3\hat{i}-7\hat{j}+6\hat{k},b_2=-3{\hat{i}}_2\hat{j}+4\hat{k}$
then, $\overrightarrow{b_1}\times \overrightarrow{b_2}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& -1& 1\\ -3& 2& 4\end{array}\right|$
$=\hat{i}(-4-2)-\hat{j}(12+3)+\hat{k}(6-3)$
$=-6\hat{i}-15\hat{j}+3\hat{k}$
$|\overrightarrow{b_1}\times \overrightarrow{b_2}|=\sqrt{(-6{)}^2+(-15{)}^2+(3{)}^2}$
$=\sqrt{36+225+9}$
$=\sqrt{270}$
And $\overrightarrow{a_2}-\overrightarrow{a_1}=(-3\hat{i}-7\hat{j}+6\hat{k})-(3\hat{i}+8\hat{j}+3\hat{k})$
$=-3\hat{i}-7\hat{j}+6\hat{k}-3\hat{i}-8\hat{j}-3\hat{k}$
$=-6\hat{i}-15\hat{j}+3\hat{k}$
Maximum Distance $=\frac{[\overrightarrow{a_2}-\overrightarrow{a_1}\cdot \overrightarrow{b_1}\overrightarrow{b_2}]}{|\overrightarrow{b_1}\times \overrightarrow{b_2}|}$
$=\frac{(\overrightarrow{a_2}-\overrightarrow{a_1})\cdot (\overrightarrow{b_1}\times \overrightarrow{b_2})}{|\overrightarrow{b_1}\times \overrightarrow{b_2}|}$
$=\frac{(-6\hat{i}-15\hat{j}+3\hat{k})(-6\hat{i}-15\hat{j}+3\hat{k})}{\sqrt{270}}$
$=\frac{36+225+9}{\sqrt{270}}$
$=\frac{270}{\sqrt{270}}$
$=\sqrt{270}$
$=3\sqrt{30}$.