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.