Q.Find the shortest distance between the lines $\bar{r} = (\hat{i}+2\hat{j}+\hat{k}) + \lambda(\hat{i}-\hat{j}+\hat{k})$ and $\bar{r} = (2\hat{i}-\hat{j}-\hat{k}) + \mu(2\hat{i}+\hat{j}+2\hat{k})$.

Main part: use the skew-lines shortest-distance formula $\dfrac{|(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)|}{|\vec b_1\times\vec b_2|}$. (OR: form the family of planes through the intersection line and impose perpendicularity.)

Main part. Line 1: $\vec a_1=\hat i+2\hat j+\hat k$, $\vec b_1=\hat i-\hat j+\hat k$. Line 2: $\vec a_2=2\hat i-\hat j-\hat k$, $\vec b_2=2\hat i+\hat j+2\hat k$.

$\vec a_2-\vec a_1 = \hat i-3\hat j-2\hat k$

$\vec b_1\times\vec b_2 = \begin{vmatrix}\hat i&\hat j&\hat k\\1&-1&1\\2&1&2\end{vmatrix} = \hat i[(-1)(2)-(1)(1)] - \hat j[(1)(2)-(1)(2)] + \hat k[(1)(1)-(-1)(2)]$

$= \hat i(-3) - \hat j(0) + \hat k(3) = -3\hat i+3\hat k$

$|\vec b_1\times\vec b_2| = \sqrt{9+0+9} = \sqrt{18} = 3\sqrt2$

$(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2) = (1)(-3)+(-3)(0)+(-2)(3) = -3+0-6 = -9$

$\text{Shortest distance} = \frac{|-9|}{3\sqrt2} = \frac{9}{3\sqrt2} = \frac{3}{\sqrt2} = \frac{3\sqrt2}{2}$

OR. Family of planes through the intersection of $x+y+z=1$ and $2x+3y+4z=5$:

$(x+y+z-1) + k(2x+3y+4z-5) = 0 \;\Rightarrow\; (1+2k)x+(1+3k)y+(1+4k)z-(1+5k)=0$ …