# Show that the lines and are coplanar and the plane containing them is given by

Given lines:

$\\ \text {As we know that two lines } \vec{r}=\vec{a}_{1}+\lambda \vec{b}_{1} \text { and } \vec{r}=\vec{a}_{2}+\mu \vec{b}_{2} \text { are coplanar if } \\ \left(\vec{a}_{2}-\vec{a}_{1}\right) \cdot\left(\vec{b}_{1} \times \vec{b}_{2}\right)=0$

$\left(\vec{a}_{2}-\vec{a}_{1}\right) =\vec{b}-\vec{a}$

$\left(\vec{b}_{1} \times \vec{b}_{2}\right)= \left(\vec{b} \times \vec{a}\right)$

$\left(\vec{b} - \vec{a}\right) \cdot \left(\vec{b} \times \vec{a}\right) =0$

Hence the two lines are coplanar.

The equation plane containing two lines is given by

$(\vec{r}-\vec{a}) .\vec{n}=0$

$\vec{n}=\left ( \overrightarrow{b} \times \overrightarrow{a} \right )$

$(\overrightarrow{r} - \overrightarrow{a} ) \cdot \left ( \overrightarrow{b} \times \overrightarrow{a} \right ) =0$

$\overrightarrow{r} \cdot \left ( \overrightarrow{b} \times \overrightarrow{a} \right ) - \overrightarrow{a} \cdot \left ( \overrightarrow{b} \times \overrightarrow{a} \right )=0$

$\because \overrightarrow{a} \cdot \left ( \overrightarrow{b} \times \overrightarrow{a} \right )=0$

The equation plane containing the given two lines $\overrightarrow{r} \cdot \left ( \overrightarrow{b} \times \overrightarrow{a} \right ) =0$

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