Vector Algebra

Can a plane be drawn through the lines $\overrightarrow{r} = \left(\hat{i} + 2 \hat{j} - 4 \hat{k}\right) + t \left(2 \hat{i} + 3 \hat{j} + 6 \hat{k}\right)$ and $\overrightarrow{r} = \left(3 \hat{i} + 3 \hat{j} - 5 \hat{k}\right) + s \left(-2 \hat{i} + 3 \hat{j} + 8 \hat{k}\right)$?

The equations of the given lines are

$L_1: \;\; \overrightarrow{r} = \left(\hat{i} + 2 \hat{j} - 4 \hat{k}\right) + t \left(2 \hat{i} + 3 \hat{j} + 6 \hat{k}\right)$

$L_2: \;\; \overrightarrow{r} = \left(3 \hat{i} + 3 \hat{j} - 5 \hat{k}\right) + s \left(-2 \hat{i} + 3 \hat{j} + 8 \hat{k}\right)$

Lines $L_1$ and $L_2$ are of the form

$\overrightarrow{r} = \overrightarrow{a_1} + t \overrightarrow{b_1}$ $\;$ and $\;$ $\overrightarrow{r}= \overrightarrow{a_2} + s \overrightarrow{b_2}$ $\;$ respectively.

Here,

$\overrightarrow{a_1} = \hat{i} + 2 \hat{j} - 4 \hat{k}$, $\;$ $\overrightarrow{b_1} = 2 \hat{i} + 3 \hat{j} + 6 \hat{k}$

$\overrightarrow{a_2} = 3 \hat{i} + 3 \hat{j} - 5 \hat{k}$, $\;$ $\overrightarrow{b_2} = -2 \hat{i} + 3 \hat{j} + 8 \hat{k}$

Now,

$\begin{aligned} \overrightarrow{a_2} - \overrightarrow{a_1} & = \left(3 \hat{i} + 3 \hat{j} - 5 \hat{k}\right) - \left(\hat{i} + 2 \hat{j} - 4 \hat{k}\right) \\\\ & = 2 \hat{i} + \hat{j} - \hat{k} \end{aligned}$

$\begin{aligned} \overrightarrow{b_1} \times \overrightarrow{b_2} & = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 6 \\ -2 & 3 & 8 \end{vmatrix} \\\\ & = \hat{i} \left(24 - 18\right) - \hat{j} \left(16 + 12\right) + \hat{k} \left(6 + 6\right) \\\\ & = 6 \hat{i} - 28 \hat{j} + 12 \hat{k} \end{aligned}$

$\begin{aligned} \left(\overrightarrow{a_2} - \overrightarrow{a_1}\right) \cdot \left(\overrightarrow{b_1} \times \overrightarrow{b_2}\right) & = \left(2 \hat{i} + \hat{j} - \hat{k}\right) \cdot \left(6 \hat{i} - 28 \hat{j} + 12 \hat{k}\right) \\\\ & = 12 - 28 - 12 = -28 \neq 0 \end{aligned}$

$\because$ $\;$ $\left(\overrightarrow{a_2} - \overrightarrow{a_1}\right) \cdot \left(\overrightarrow{b_1} \times \overrightarrow{b_2}\right) \neq 0$ $\implies$ lines $L_1$ and $L_2$ are not coplanar.

$\because$ $\;$ The given lines are not coplanar, a plane cannot be drawn through the lines $L_1$ and $L_2$.