(a) Let
\(\vec {a_1} = 0\)
\(\vec{a_2} = 3\hat i + 3 \hat j \)
\(\vec {b_1} = \hat i + 2\hat j- \hat k\)
\(\vec {b_2} =2\hat i + \hat j + \hat k \)
\(\vec {a_2 } - \vec{a_1} = 3\hat i + 3\hat j\)
\(\vec {b_1} \times \vec{b_2} = \begin{vmatrix}\hat i&\hat j&\hat k\\1&2&-1\\2&1&1\end{vmatrix}\)
\(= 3\hat i - 3\hat j - 3\hat k\)
\((\vec{a_1}- \vec{a_1}) .(\vec{b_1}\times\vec{b_2}) = (3\hat i + 3 \hat j). ( 3\hat i - 3\hat j - 3\hat k)\)
\(= 9 - 9\)
= 0
\(\therefore\) Shortest distance between line is
\(d = \left|\frac{(\vec{a_2}- \vec{a_1}).(\vec{b_1}\times \vec{b_2})}{|\vec{b_1} \times \vec{b_2}|}\right| = \frac{0}{3\sqrt3}= 0\,units\)
(b) From equations of both lines, we get
\(\lambda (\hat i +2\hat j- \hat k)= (3\hat i+ 3\hat j)+ \mu(2\hat i + \hat j ++ \hat k)\)
(\(\because\) If they collide then they meet each other at that point)
⇒ \(\lambda \hat i + 2 \lambda \hat j - \lambda\hat k =(3 + 2\mu)\hat i + ( 3 + \mu)\hat j + \mu \hat k\)
\(\lambda = 3 + 2 \mu \) ......(1) (By comparing component of \(\hat i\))
\(2 \mu = 3 + \mu\) ......(2) (By comparing component of \(\hat j\))
\(-\lambda = \mu\) ......(3) (By comparing component of \(\hat k\))
From (3) & (1), we obtain
\(\lambda = 3 - 2\lambda\)
⇒ \(3 \lambda = 3\)
⇒ \(\lambda = 1\)
From (3) & (2), also we obtain
\(2\lambda = 3 - \lambda\)
⇒ \(3\lambda = 3\)
⇒ \(\lambda = 1\)
Hence, \(\lambda = 1\) & \(\mu = -1\) gives the collidel point.
Put \(\lambda = 1\) in equation of line (1), we obtain
\(\vec r = \hat i + 2\hat j - \hat k\)
Hence at point \(x = 1, y = 2 \) & \(z = -1 \) or (1, 2, -1) both motorcycles collide.
