Question

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

Hint:

Use d = |(a2-a1).(b1 x b2)| / |b1 x b2| for two skew lines in vector form.

Answer 1

Step 1: Extract position and direction vectors directly from the vector equations:
Line 1 passes through $(1,2,1)$ along $(1,-1,1)$; line 2 passes through $(2,-1,-1)$ along $(2,1,2)$.

Step 2: Form the connecting vector between the two base points:
$\vec a_2-\vec a_1=(2-1,-1-2,-1-1)=(1,-3,-2)$.

Step 3: Cross the two direction vectors to get the common perpendicular direction:
$(1,-1,1)\times(2,1,2)$: $i$-component $=(-1)(2)-(1)(1)=-3$; $j$-component $=-[(1)(2)-(1)(2)]=0$; $k$-component $=(1)(1)-(-1)(2)=3$. Result $(-3,0,3)$, magnitude $3\sqrt2$.

Step 4: Project the connecting vector onto this common perpendicular:
Dot product $(1,-3,-2)\cdot(-3,0,3)=-3+0-6=-9$; distance is $|{-9}|/(3\sqrt2)=3/\sqrt2$.

Final Answer:
\[ \boxed{d=3\sqrt2/2} \]