Can u draw a plane through the given two lines? justify your answer.$\overrightarrow{r}(\overrightarrow{i}+\overrightarrow{2j}-\overrightarrow{4k})+t (\overrightarrow{2i}+\overrightarrow{3j} +\overrightarrow{6k})$ and $r=(\overrightarrow{3i}+\overrightarrow{3j}-\overrightarrow{5k})+ s(-\overrightarrow{2i}+\overrightarrow{3j}+\overrightarrow{8k})$

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• The lines $\overrightarrow r=\overrightarrow a_1+t\overrightarrow u, \: \overrightarrow r=\overrightarrow a_2+s\overrightarrow u$ intersect if $[(\overrightarrow a_2-\overrightarrow a_1)\overrightarrow u\: \overrightarrow v]=0$
Step 1
A plane can be drawn through the lines
$\overrightarrow r=(\overrightarrow i+2\overrightarrow j-4\overrightarrow k)+t(2\overrightarrow i+3\overrightarrow j+6\overrightarrow k)$ (i) and
$\overrightarrow r = (3\overrightarrow i+3\overrightarrow j-5\overrightarrow k)+s(-2\overrightarrow i+3\overrightarrow j+8\overrightarrow k)$ (ii)
Provided the two lines are not skew lines.
(i) is parallel to $\overrightarrow u=2\overrightarrow i+3\overrightarrow j+6\overrightarrow k$ and passes through the point whose pv is $\overrightarrow a_1=\overrightarrow i+2\overrightarrow j-4\overrightarrow k$
(ii) is parallel to $\overrightarrow v = -2\overrightarrow i+3\overrightarrow j+8\overrightarrow k$ and passes through the point whose pv is $\overrightarrow a_2 = 3\overrightarrow i+3\overrightarrow j-5\overrightarrow k$
The lines are not parallel. The lines are intersecting ( not skew lines ) when $[\overrightarrow a_2-\overrightarrow a_1\: \overrightarrow u\: \overrightarrow v]=0$ since these 3 vectors are coplanar.
Step 2
$\overrightarrow a_2-\overrightarrow a_1=(3-1)\overrightarrow i+(3-2)\overrightarrow j+(-5+4)\overrightarrow k = 2\overrightarrow i+\overrightarrow j-\overrightarrow k$
$[ \overrightarrow a_2-\overrightarrow a_1\: \overrightarrow u\: \overrightarrow v]= \begin{vmatrix} 2 & 1 & -1 \\ 2 & 3 & 6 \\ -2 & 3 & 8 \end{vmatrix}$
$= 2 (24-18)-1(16+12)-1(6+6)$
$= 12-28-12=-28 \neq 0$
Step 3
Since $[\overrightarrow a_2-\overrightarrow a_1\: \overrightarrow u\: \overrightarrow v] \neq 0,$ the lines are skew.
$\therefore$ a plane cannot be drawn through them.