Q.Show that the line $\vec r=(\hat i+2\hat j+3\hat k)+\lambda(\hat i+\hat j+\hat k)$ and the line $\vec r=(-\hat j+2\hat k)+\mu(2\hat i+\hat j-\hat k)$ are skew, and find the shortest distance between them.

Here $\vec a_1=\hat i+2\hat j+3\hat k$, $\vec b_1=\hat i+\hat j+\hat k$; $\vec a_2=-\hat j+2\hat k$, $\vec b_2=2\hat i+\hat j-\hat k$.

Not parallel: $(1,1,1)$ is not a scalar multiple of $(2,1,-1)$.

Not intersecting: equating components, $1+t=2s,\ 2+t=-1+s,\ 3+t=2-s$. From the second, $t=s-3$; substituting into the first, $1+(s-3)=2s \Rightarrow s=-2,\ t=-5$. Checking the third: $3+(-5)=-2$ but $2-(-2)=4$ -- these disagree, so no common point exists. Hence the lines are skew.

Shortest distance: $\vec b_1\times\vec b_2=\begin{vmatrix}\hat i&\hat j&\hat k\\1&1&1\\2&1&-1\end{vmatrix}=\hat i(-1-1)-\hat j(-1-2)+\hat k(1-2)=-2\hat i+3\hat j-\hat k$, with $|\vec b_1\times\vec b_2|=\sqrt{4+9+1}=\sqrt{14}$. …

First verify the two lines are neither parallel nor intersecting (establishing they are skew); then compute $\vec b_1\times\vec b_2$, the vector joining the …

Sign errors expanding the $2\times2$ minors of the cross-product determinant (the most common real defect in this topic); forgetting the absolute value and reporting a negative distanc …