Question

Find the shortest distance between lines $\overrightarrow{r}$=$6\hat i+2\hat j+2\hat k$+λ($\hat i+2\hat j+2\hat k$)and$\overrightarrow{r}$=-$-4\hat i-\hat k$+μ($3\hat i+2\hat j+2\hat k$).

Answer 1

The given lines are
$\overrightarrow{r}$=$6\hat i+2\hat j+2\hat k$+λ($\hat i+2\hat j+2\hat k$)...(1)
$\overrightarrow{r}$=$-4\hat i-\hat k$+μ($3\hat i+2\hat j+2\hat k$)...(2)

It is known that the shortest distance between two lines,$\overrightarrow{r}$=$\overrightarrow{a_1}$+λ$\overrightarrow{b_1}$and $\overrightarrow{r}$=$\overrightarrow{a_2}$+λ$\overrightarrow{b_2}$, is given by
d=|($\overrightarrow{b_1}$×$\overrightarrow{b_2}$).($\overrightarrow{a_2}$-$\overrightarrow{a_1}$) / |$\overrightarrow{b_1}$×$\overrightarrow{b_2}$||...(3)

Comparing $\overrightarrow{r}$=$\overrightarrow{a_1}$→+λ$\overrightarrow{b_1}$→ and $\overrightarrow{r}$=$\overrightarrow{a_2}$+λ$\overrightarrow{b_2}$→ to equations(1) and (2), we obtain

$\overrightarrow{a_1}$=$6\hat i+2\hat j+2\hat k$ $\overrightarrow{b_1}$→=$\hat i+2\hat j+2\hat k$ $\overrightarrow{a_2}$=-$-4\hat i-\hat k$ $\overrightarrow{b_2}$ = $3\hat i+2\hat j+2\hat k$

⇒$\overrightarrow{a_1}$-$\overrightarrow{a_2}$=($-4\hat i-\hat k$)-($6\hat i+2\hat j+2\hat k$)=$-10\hat i-2\hat j-3\hat k$

⇒$\overrightarrow{b_1}$×$\overrightarrow{b_2}$=$\begin{vmatrix} \hat i & \hat j & \hat k\\\ 1 & -2 & 2 \\\ 3 &-2&-2\end{vmatrix}$=(4+4)$\hat i$-(-2-6)$\hat k$=$8\hat i+8\hat j+4\hat k$

∴|$\overrightarrow{b_1}$×$\overrightarrow{b_2}$|

=$\sqrt{√(8)^2+(8)^2+(4)^2}$

=12 ($\overrightarrow{b_1}$×$\overrightarrow{b_2}$).(a2-a1)
=($8\hat i+8\hat j+4\hat k$).($-10\hat i-2\hat j-3\hat k$)
=-80-16-12
=-108

Substituting all the values in equation(1), we obtain
d=|-108/12|=9

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