Find the shortest distance between the lines whose vector equations are : vector r = (1 - t)hat i + (t - 2 )hat j + (3 - 2t) hat k,)

Question

Find the shortest distance between the lines whose vector equations are :

\(\vec r\)= (1 − t)\(\hat i\)+ (t − 2)\(\hat j\)+ (3 − 2)\(\hat k\), \(\vec r\)= (s + 1)\(\hat i\)+ (2 − 1)\(\hat j\)− (2 + 1)\(\hat k\).

vector r = (1 - t) i + (t - 2 )j + (3 - 2t)k, vector r = (s +1)i + (2 - 1)j - (2 - 1)k.

Answer 1

Given vector equations of both lines are

\(\vec r\)= (1 − t)\(\hat i\)+ (t − 2)\(\hat j\)+ (3 − 2t)\(\hat k\), 

\(\vec r\)= (s + 1)\(\hat i\)+ (2s − 1)\(\hat j\)− (2s + 1)\(\hat k\). 

⇒ \(\vec r\)= (\(\hat i\)− 2\(\hat j\)+ 3\(\hat k\)) + (−\(\hat i\)+ \(\hat j\)− 2\(\hat k\)) = \(\overset\longrightarrow {a_1}\)+ t\(\overset\longrightarrow {b_1}\), 

\(\vec r\)= (\(\hat i\)− \(\hat j\)− \(\hat k\)) + s(\(\hat i\)+ 2\(\hat j\)− 2\(\hat k\)) = \(\overset\longrightarrow {a_2}\)+ s\(\overset\longrightarrow {b_2}\).

Therefore, shortest distance between both lines is

Hence, the shortest distance between the given line is d = \(\left|\frac{(\overset\longrightarrow {b_1} \times {\overset\longrightarrow{b_2}})(\overset\longrightarrow {a_2}\times\overset\longrightarrow{a_1})}{|\overset\longrightarrow{b_1} \times \overset\longrightarrow{b_2}|}\right|\) = \(|\frac{8}{\sqrt{29}}|\) = \(\frac{8}{\sqrt{29}}\) unit.