Show that the lines r =(2 i -3 k )+ ( i +2 j +3 k ) and r =(2 i +6 j +3 k )+ (2 i +3 j +4 k ) are coplanar. Also, find the equation of the plane containing them.

We know that the lines r = a _1+ b _1 and r = a _2+ b _2 are coplanar if a _1 ( b _1 b _2)= a _2 ( b _1 b _2) and the equation of the plane containing them is r ( b _1 b _2)= a _1 ( b _1 b _2) Here, a _1=0 i +2 j -3 k , b _1= i +2 j +3 k a _2=2 i +6 j +3 k , b _2=2 i +3 j +4 k b _1 b _2= i & j & k 1&2&3 2&3&4 =- i +2 j - k a _1 ( b _1 b _2)=(0 i +2 j -3 k ) (- i +2 j - k ) =0+4+3=7 a _2 ( b _1 b _2)=(2 i +6 j +3 k ) (- i +2 j - k ) =-2+12-3=7 Clearly, a _1 ( b _1 b _2)= a _2 ( b _1 b _2) Hence, the given lines are coplanar. The equation of the plane containing the given lines is r ( b _1 b _2)= a _1 ( b _1 b _2) r (- i +2 j - k )=(0 i +2 j -3 k ) (- i +2 j - k ) r (- i +2 j - k )=7 r (- i +2 j - k )+7=0

Show that the lines r =(2 i -3 k )+ ( i +2 j +3 k ) and r =(2 i +…

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Question

Show that the lines $\vec{\text{r}}=(2\hat{\text{i}}-3\hat{\text{k}})+\lambda(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}})$ and $\vec{\text{r}}=(2\hat{\text{i}}+6\hat{\text{j}}+3\hat{\text{k}})+\mu(2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}})$ are coplanar. Also, find the equation of the plane containing them.

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Answer

We know that the lines $\vec{\text{r}}=\vec{\text{a}}_1+\lambda\vec{\text{b}}_1$ and $\vec{\text{r}}=\vec{\text{a}}_2+\mu\vec{\text{b}}_2$ are coplanar if
$\vec{\text{a}}_1\cdot(\vec{\text{b}}_1\times\vec{\text{b}}_2)=\vec{\text{a}}_2\cdot(\vec{\text{b}}_1\times\vec{\text{b}}_2)$ and the equation of the plane containing them is
$\vec{\text{r}}\cdot(\vec{\text{b}}_1\times\vec{\text{b}}_2)=\vec{\text{a}}_1\cdot(\vec{\text{b}}_1\times\vec{\text{b}}_2)$
Here,
$\vec{\text{a}}_1=0\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}},\vec{\text{b}}_1=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$
$\vec{\text{a}}_2=2\hat{\text{i}}+6\hat{\text{j}}+3\hat{\text{k}},\vec{\text{b}}_2=2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}}$
$\vec{\text{b}}_1\times\vec{\text{b}}_2=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&2&3\\2&3&4\end{vmatrix}$
$=-\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}$
$\vec{\text{a}}_1\cdot(\vec{\text{b}}_1\times\vec{\text{b}}_2)=(0\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}})\cdot(-\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}})$
$=0+4+3=7$
$\vec{\text{a}}_2\cdot(\vec{\text{b}}_1\times\vec{\text{b}}_2)=(2\hat{\text{i}}+6\hat{\text{j}}+3\hat{\text{k}})\cdot(-\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}})$
$=-2+12-3=7$
Clearly, $\vec{\text{a}}_1\cdot(\vec{\text{b}}_1\times\vec{\text{b}}_2)=\vec{\text{a}}_2\cdot(\vec{\text{b}}_1\times\vec{\text{b}}_2)$
Hence, the given lines are coplanar.
The equation of the plane containing the given lines is
$\vec{\text{r}}\cdot(\vec{\text{b}}_1\times\vec{\text{b}}_2)=\vec{\text{a}}_1\cdot(\vec{\text{b}}_1\times\vec{\text{b}}_2)$
$\Rightarrow\vec{\text{r}}\cdot(-\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}})=(0\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}})\cdot(-\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}})$
$\Rightarrow\vec{\text{r}}\cdot(-\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}})=7$
$\Rightarrow\vec{\text{r}}\cdot(-\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}})+7=0$

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