Question

Find the shortest distance between the skew lines $r=(6i+2j+2k)+t(i-2j+2k)$ and $F=(-4i-k)+s(3i-2j-2k)$ where s,t are scalars.

Answer 1

shortest distance between lines vector equations

$\overrightarrow{r}=\overrightarrow{a_1}+t\overrightarrow{b_1}$ and $\overrightarrow{r}=\overrightarrow{a_2}+s\overrightarrow{b_2}$ is

$\left|\frac{(\overrightarrow{b_1}\times \overrightarrow{b_2}).(\overrightarrow{a_2}-\overrightarrow{a_1})}{|\overrightarrow{b_1}\times \overrightarrow{b_2}|}\right|$

Now,

$\overrightarrow{r}=(6i+2j+2k)+t(i-2j+2k)$

Comparing with

$\overrightarrow{r}=\overrightarrow{a_1}+t\overrightarrow{b_1}$

$\overrightarrow{a_1}=6i+2j+2k$

$\overrightarrow{b_1}=i-2j+2k$

$\overrightarrow{r}=(-4i-k)+s(3i-2j-2k)$

Comparing with

$\overrightarrow{r}=\overrightarrow{a_2}+s\overrightarrow{b_2}$

$\overrightarrow{a_2}=-4i-k$

$\overrightarrow{b_2}=3i-2j-2k$

Now,

$(\overrightarrow{a_2}-\overrightarrow{a_1})=(-4i-k)-(6i+2j+2k)$

$=(-4-6)i+(-2)j+(-1-2)k$

$=-10i-2j-3k$

$(\overrightarrow{b_1}\times \overrightarrow{b_2})=\left|\begin{array}{ccc}i& j& k\\ 1& -2& 2\\ 3& -2& -2\end{array}\right|$

$=i\left[\right(-2\times -2)-(-2\times 2\left)\right]-j\left[\right(1\times -1)-(3\times 2\left)\right]+k\left[\right(1\times -2)-(3\times -2\left)\right]$

$=i[4+4]-j[-2-6]+k[-2+6]$

$=8i+8j+4k$

$|\overrightarrow{b_1}\times \overrightarrow{b_2}|=\sqrt{8^2+8^2+4^2}$

$=\sqrt{64+64+16}$

$=\sqrt{144}$

$=12$

Also,

$(\overrightarrow{b_1}\times \overrightarrow{b_2}).(\overrightarrow{a_2}-\overrightarrow{a_1})=(8i+8j+4k).(-10i-2j-3k)$

$=(8\times -10)+(8\times -2)+(4\times -3)$

$=-80-16-12$

$=-108$

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

$=\left|\frac{-108}{12}\right|$

$=|-9|$

$=9$

Therefore, the shortest distance between the given two lines is 9.