Let $P$ and $Q$ be points on the lines
$L_1:\vec{r}=6\hat{i}+7\hat{j}+4\hat{k}+\lambda(3\hat{i}-1\hat{j}+\hat{k})$ and $L_2:\vec{r}=-9\hat{j}+2\hat{k}+\mu(-3\hat{i}+2\hat{j}+4\hat{k})$ respectively which are nearest to each other.If $O$ is the origin.Then find $\vec{OP},\vec{OQ}$ i.e. the position vectors of $P$ and $Q.$
The given two lines are skew lines as they are neither parallel nor intersecting lines.
As $P$ is a point on $L_1:\vec{r}=6\hat{i}+7\hat{j}+4\hat{k}+\lambda(3\hat{i}-1\hat{j}+\hat{k})$,so its coordinates are $(3\lambda+6,-\lambda+7,\lambda+4)$
As $Q$ is a point on $L_2:\vec{r}=-9\hat{j}+2\hat{k}+\mu(-3\hat{i}+2\hat{j}+4\hat{k})$,so its coordinates are $(-3\mu,2\mu-9,4\mu+2)$
Let distance $PQ=d$
$d^2=(3\lambda+3\mu+6)^2+(-\lambda+7-2\mu+9)^2+(\lambda+4-4\mu-2)^2$
As $d$ is the shortest distance,I need to use calculus to find the minimum value of $d^2$.As there are two variables $\lambda$ and $\mu$ in the expression for $d^2$,so i used partial differentiation.
Put $\frac{\partial(d^2)}{\partial \lambda}=0$gives $11\lambda+7\mu+4=0$
Put $\frac{\partial(d^2)}{\partial \mu}=0$gives$7\lambda+29\mu=22$
Solving the two equations,i get $\lambda=-1,\mu=1$
$P(3\lambda+6,-\lambda+7,\lambda+4)=(3,8,3)$ and $Q(-3\mu,2\mu-9,4\mu+2)=(-3,-7,6)$
So $\vec{OP}=3\hat{i}+8\hat{j}+3\hat{k}$
$\vec{OQ}=-3\hat{i}-7\hat{j}+6\hat{k}$
This is a long and tedious method of finding the points of shortest distance.
I want to ask that is there any other method possible which is short and elegant
than this one.Such question comes in competition exam in objective questions,where there is limited time to solve them.
Please help me.