Question

Find the shortest distance between the lines whose vector equations are

\(\overrightarrow r=(1-t)\hat i+(t-2)\hat j+(3-2t)\hat k\) and

\(\overrightarrow r=(s+1)\hat i+(2s-1)\hat j-(2s+1)\hat k\)

Answer 1

The given lines are

\(\overrightarrow r=(1-t)\hat i+(t-2)\hat j+(3-2t)\hat k\)
\(\Rightarrow\overrightarrow r=(\hat i-2\hat j+3\hat k)+t(-\hat i+\hat j-2\hat k)\)...(1)

\(\overrightarrow r=(s+1)\hat i+(2s-1)\hat j-(2s+1)\hat k\)
\(\Rightarrow\overrightarrow r=(\hat i-\hat j+\hat k)+s(\hat i+2\hat j-2\hat k)\)...(2)

It is known that the shortest distance between the lines,
\(\overrightarrow r=\overrightarrow a_1+\lambda \overrightarrow b_1\) and \(\overrightarrow r=\overrightarrow a_2+\mu \overrightarrow b_2\), is given by,
d=|(b1→×b2→).(a2→-a2→)/|b1→×b2→||...(3)

For the given equations,
\(\overrightarrow a_1= \hat i-2\hat j+3\hat k\)
\(\overrightarrow b_1= -\hat i+\hat j-2\hat k\)
\(\overrightarrow a_2= \hat i-\hat j-\hat k\)
\(\overrightarrow b_2= \hat i+2\hat j-2\hat k\)

\(\overrightarrow a_2-\overrightarrow a_1\)
=\((\hat i-\hat j-\hat k)\)-\((\hat i-2\hat j+3\hat k)\)=\(\hat j-4\hat k\) 

\(\overrightarrow b_1.\overrightarrow b_2\) = \(\begin{vmatrix}\hat i&\hat j&\hat k\\-1&1&-2\\1&2&-2\end{vmatrix}\)

=\((-2+4)\hat i-(2+2)\hat j+(-2-1)\hat k\)
=\(2\hat i-4\hat j-3\hat k\)

\(\Rightarrow \mid \overrightarrow b_1*\overrightarrow b_2\mid\)
=\(\sqrt{(2)^2+(-4)^2+(-3)^2}\)
=\(\sqrt{4+16+9}\)
=\(\sqrt{29}\)

∴(\( \overrightarrow b_1*\overrightarrow b_2\)).(\(\overrightarrow a_2-\overrightarrow a_1\))
=(\(2\hat i-4\hat j-3\hat k\)).(\(\hat j-4\hat k\)^)
=-4+12
=8

Substituting all the values in equation (3), we obtain
d=|\(\frac{8}{\sqrt{29}}\)|
=\(\frac{8}{\sqrt{29}}\)

Therefore, the shortest distance between the lines is \(\frac{8}{\sqrt{29}}\) units.