$\overrightarrow {{\rm{AB}}} = 3\widehat {\rm{i}} - \widehat {\rm{j}} + \widehat {\rm{k}}$

and $\overrightarrow {{\rm{CD}}} = - 3\widehat {\rm{i}} + 2\widehat {\rm{j}} + 4\widehat {\rm{k}}$ are two vectors.z The position vectors of the points $A$ and $C$ are $6\widehat {\rm{i}} + 7\widehat {\rm{j}} + 4\widehat {\rm{k}}$ and $- 9\widehat {\rm{i}} + 2\widehat {\rm{k}}$, respectively. Find the position vector of a point $P$ on the line $A B$ and a point $Q$ on the line $CD$ such that $\overrightarrow {{\rm{PQ}}}$ is perpendicular to $\overrightarrow {{\rm{AB}}}$ and $\overrightarrow {{\rm{CD}}}$ both.

It is given that, $\overrightarrow {{\rm{AB}}} = 3\widehat {\rm{i}} - \widehat {\rm{j}} + \widehat {\rm{k}}$

and $\overrightarrow {{\rm{CD}}} = - 3\widehat {\rm{i}} + 2\widehat {\rm{j}} + 4\widehat {\rm{k}}$

Also, the position vectors of $A$ and $C$ are $6\widehat {\rm{i}} + 7\widehat {\rm{j}} + 4\widehat {\rm{k}}$ and $- 9\widehat {\rm{j}} + 2\widehat {\rm{k}}$,

respectively. Since, $\overrightarrow {{\rm{PQ}}}$ is perpendicular to both $\overrightarrow {AB}$ and $\overrightarrow {CD}$.

So, $P$ and $Q$ will be foot of perpendicular to both the lines through $A$ and $C$.

Now, equation of the line through ${\rm{A}}$ and parallel to the vector $\overrightarrow {{\rm{AB}}}$ is,

$\overrightarrow {\rm{r}} = (6\widehat {\rm{i}} + 7\widehat {\rm{j}} + 4\widehat {\rm{k}}) + \lambda (3\widehat {\rm{i}} - \widehat {\rm{j}} + \widehat {\rm{k}})$

and the line through $C$ and parallel to the vector $\overrightarrow {{\rm{CD}}}$ is

given by $\overrightarrow {\rm{r}} = - 9\widehat {\rm{j}} + 2\widehat {\rm{k}} + \mu ( - 3\widehat {\rm{i}} + 2\widehat {\rm{j}} + 4\widehat {\rm{k}})$

……(i)
Let $\overrightarrow {\rm{r}} = (6\widehat {\rm{i}} + 7\widehat {\rm{j}} + 4\widehat {\rm{k}}) + \lambda (3\widehat {\rm{i}} - \widehat {\rm{j}} + \widehat {\rm{k}})$

and $\overrightarrow {\rm{r}} = - 9\widehat {\rm{j}} + 2\widehat {\rm{k}} + \mu ( - 3\widehat {\rm{i}} + 2\widehat {\rm{j}} + 4\widehat {\rm{k}})$

……(ii)
Let $P(6 + 3\lambda ,7 - \lambda ,4 + \lambda )$ is any point on the first line and $Q$ be any point on second line is given by $( - 3\mu , - 9 + 2\mu ,2 + 4\mu )$.

$\therefore \overrightarrow {{\rm{PQ}}} = ( - 3\mu - 6 - 3\lambda )\widehat {\rm{i}} + ( - 9 + 2\mu - 7 + \lambda )\widehat {\rm{j}} + (2 + 4\mu - 4 - \lambda )\widehat {\rm{k}}$

$= ( - 3\mu - 6 - 3\lambda )\widehat {\rm{i}} + (2\mu + \lambda - 16)\widehat {\rm{j}} + (4\mu - \lambda - 2)\widehat {\rm{k}}$

If $\overrightarrow {{\rm{PQ}}}$ is perpendicular to the first line, then $3( - 3\mu - 6 - 3\lambda ) - (2\mu + \lambda - 16) + (4\mu - \lambda - 2) = 0$
$\Rightarrow$

$\Rightarrow$ $- 7\mu - 11\lambda - 4 = 0$

……..(iii)
If ${\rm{PQ}}$ is perpendicular to the second line, then $- 3( - 3\mu - 6 - 3\lambda ) + 2(2\mu + \lambda - 16) + 4(4\mu - \lambda - 2) = 0$

$\Rightarrow$ $9\mu + 18 + 9\lambda + 4\mu + 2\lambda - 32 + 16\mu - 4\lambda - 8 = 0$

$\Rightarrow$ $29\mu + 7\lambda - 22 = 0$

……(iv)
On solving Eqs. (iii) and (iv),

we get
$- 49\mu - 77\lambda - 28 = 0$
$\Rightarrow$ $319\mu + 77\lambda - 242 = 0$

$\Rightarrow$ $270\mu - 270 = 0$

$\Rightarrow$ $\mu = 1$
Using $\mu$ in Eq. (iii),

we get
$- 7(1) - 11\lambda - 4 = 0$
$\Rightarrow$ $- 7 - 11\lambda - 4 = 0$
$\Rightarrow$ $- 11 - 11\lambda = 0$
$\Rightarrow$ $\lambda = - 1$

$\therefore \overrightarrow {PQ} = [ - 3(1) - 6 - 3( - 1)]\widehat {\rm{i}} + [2(1) + ( - 1) - 16]\widehat {\rm{j}} + [4(1) - ( - 1) - 2]\widehat {\rm{k}}$

$= - 6\widehat {\rm{i}} - 15\widehat {\rm{j}} + 3\widehat {\rm{k}}$