Question

Define skew lines. Using only vector approach, find the shortest distance between the following two skew lines:
$\overrightarrow{r}=(8+3\lambda )\hat{i}-(9+16\lambda )\hat{j}+(10+7\lambda )\hat{k}$
$\overrightarrow{r}=15\hat{i}+29\hat{j}+5\hat{k}+\mu (3\hat{i}+8\hat{j}-5\hat{k})$

Answer 1

The given equations are

$\overrightarrow{r}=8\hat{i}-9\hat{j}+10\hat{k}+\mu (3\hat{i}-16\hat{j}+7\hat{k})......\left(i\right)$

$\overrightarrow{r}=15\hat{i}+29\hat{j}+5\hat{k}+\mu (3\hat{i}+8\hat{j}-7\hat{k})......\left(ii\right)$

Here,

$\overrightarrow{a_1}=8\hat{i}-9\hat{j}+10\hat{k}$

$\overrightarrow{a_2}=15\hat{i}+29\hat{j}+5\hat{k}$

$\overrightarrow{b_1}=3\hat{i}-16\hat{j}+7\hat{k}$

$\overrightarrow{b_2}=3\hat{i}+8\hat{j}-5\hat{k}$

Now, $\overrightarrow{a_2}-\overrightarrow{a_1}=(15-8)\hat{i}+(29+9)\hat{j}+(5-10)\hat{k}$

$=7\hat{i}+38\hat{j}-5\hat{k}$

And $\overrightarrow{b_1}\times \overrightarrow{b_2}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& -16& 7\\ 3& 8& -5\end{array}\right|$

$=24\hat{i}+36\hat{j}+72\hat{k}$

$(\overrightarrow{b_1}\times \overrightarrow{b_2})\cdot (\overrightarrow{a_2}-\overrightarrow{a_1})=(24\hat{i}+36\hat{j}+72\hat{k})\cdot (7\hat{i}+38\hat{j}-5\hat{k})$

$=168+1368-360$

$=1176$

Shortest distance $=\left|\frac{(\overrightarrow{b_1}\times \overrightarrow{b_2})\cdot (\overrightarrow{a_2}-\overrightarrow{a_1})}{|\overrightarrow{b_1}\times \overrightarrow{b_2}|}\right|$

$=\left|\frac{1176}{\sqrt{{24}^2+{36}^2+{72}^2}}\right|$

$=\frac{1176}{\sqrt{7056}}$

$=\frac{1176}{84}$

$=14unit$