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Show that the lines \overrightarrow{r} = \overrightarrow{a} + \lambda \overrightarrow{b} and \overrightarrow{r} = \overrightarrow{b} + \mu \overrightarrow{a} are coplanar and the plane containing them is given by \overrightarrow{r} .\left ( \overrightarrow{a} \times \overrightarrow{b} \right ) =0

 

 

Answers (1)
S safeer

Given lines:

\overrightarrow{r} = \overrightarrow{a} + \lambda \overrightarrow{b}

\overrightarrow{r} = \overrightarrow{b} + \mu \overrightarrow{a}

\\ \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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