Question

Find the shortest distance between the lines \(\vec r=\hat i+\hat j-\lambda(2\hat i+\hat j+\hat k)\) and \(\vec r=2\hat i+\hat j+\hat k+\mu(3\hat i-5\hat j+2\hat k)\).

Hint:

Use the shortest-distance-between-skew-lines formula with the cross product of the direction vectors.

Answer 1

Step 1: Key Formula or Approach:
For two skew lines \(\vec r=\vec a_1+\lambda\vec d_1\) and \(\vec r=\vec a_2+\mu\vec d_2\), the shortest distance is \(\left|\dfrac{(\vec a_2-\vec a_1)\cdot(\vec d_1\times\vec d_2)}{|\vec d_1\times\vec d_2|}\right|\).

Step 2: Reading off the data:
\(\vec a_1=\hat i+\hat j\), \(\vec d_1=2\hat i+\hat j+\hat k\) (the minus sign just flips the direction, which does not change the line or the distance); \(\vec a_2=2\hat i+\hat j+\hat k\), \(\vec d_2=3\hat i-5\hat j+2\hat k\).

Step 3: Computing \(\vec d_1\times\vec d_2\):
\[ \vec d_1\times\vec d_2 = \begin{vmatrix}\hat i&\hat j&\hat k\\2&1&1\\3&-5&2\end{vmatrix} = \hat i(1\cdot2-1\cdot(-5)) - \hat j(2\cdot2-1\cdot3) + \hat k(2\cdot(-5)-1\cdot3) \]
\[ = \hat i(2+5) - \hat j(4-3) + \hat k(-10-3) = 7\hat i-\hat j-13\hat k \]

Step 4: Finding \(|\vec d_1\times\vec d_2|\) and \((\vec a_2-\vec a_1)\):
\[ |\vec d_1\times\vec d_2| = \sqrt{49+1+169} = \sqrt{219} \]
\[ \vec a_2-\vec a_1 = (2-1)\hat i+(1-1)\hat j+(1-0)\hat k = \hat i+\hat k \]

Step 5: Dotting and dividing:
\[ (\vec a_2-\vec a_1)\cdot(\vec d_1\times\vec d_2) = (1)(7)+(0)(-1)+(1)(-13) = 7-13 = -6 \]
\[ \text{Distance} = \left|\frac{-6}{\sqrt{219}}\right| = \frac{6}{\sqrt{219}} \]

Final Answer:
The shortest distance is \(\dfrac{6}{\sqrt{219}}\) units. \[ \boxed{\dfrac{6}{\sqrt{219}} \text{ units}} \]