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# Question and Answer

## QuestionMathsClass 12

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

## Answer

Given lines are $$\vec { r } = ( \hat { i } + \hat { j } - \hat { k } ) + \lambda ( { 3 } \hat { i } - \hat { j } )$$
and $$\vec { r } = ( 4 \hat { i } - \hat { k } ) + \mu ( 2 \hat { i } + 3 \hat { k } )$$
On comparing both equations of lines with
$$\vec { r } = \vec { a } + \lambda \vec { b }$$ respectively, we get ,
$$\vec { a _ { 1 } } = \hat { i } + \hat { j } - \hat { k } , \vec { b _ { 1 } } = { 3 } \hat { i } - \hat { j }$$
and $$\vec { a _ { 2 } } = 4 \hat { i } - \hat { k } , \vec { b _ { 2 } } = 2 \hat { i } + 3 \hat { k }$$
Now $$\vec { b _ { 1 } } \times \vec { b _ { 2 } } = \left| \begin{array} { r r r } { \hat { i } } & { \hat { j } } & { \hat { k } } \\ { 3 } & { - 1 } & { 0 } \\ { 2 } & { 0 } & { 3 } \end{array} \right|$$
$$= \hat { i } ( - 3 - 0 - \hat { j } ( 9 - 0 ) + \hat { k } ( 0 + 2 )$$
$$= - 3 \hat { i } - 9 \hat { j } + 2 \hat { k }$$
and $$\vec { a _ { 2 } } - \vec { a _ { 1 } } = \left({ 4 \hat i } _ { } - \hat { k } \right) - { ( \hat i } + \hat { j } - \hat { k } ) = 3 \hat { i} - \hat { j }$$
Now,  $$\left( {{{\vec a}_2} - {{\vec a}_1}} \right)\cdot\left( {{{\vec b}_1} \times {{\vec b}_2}} \right) = (3\hat i- \hat j )$$.$$( - 3 \hat { i } - 9 \hat { j } + 2 \hat { k } )$$
= - 9 + 9 = 0
Hence, given lines are coplanar.
Now, cartesian equations of given lines are
$$\frac { x - 1 } { 3 } = \frac { y - 1 } { - 1 } = \frac { z + 1 } { 0 }$$
and $$\frac { x - 4 } { 2 } = \frac { y - 0 } { 0 } = \frac { z + 1 } { 3 }$$
Then, equation of plane containing them is
$$\left| \begin{array} { c c c } { x - x _ { 1 } } & { y - y _ { 1 } } & { z - z _ { 1 } } \\ { a _ { 1 } } & { b _ { 1 } } & { c _ { 1 } } \\ { a _ { 2 } } & { b _ { 2 } } & { c _ { 2 } } \end{array} \right| = 0$$
$$\Rightarrow \left| \begin{array} { c c c } { x - 1 } & { y - 1 } & { z + 1 } \\ { 3 } & { - 1 } & { 0 } \\ { 2 } & { 0 } & { 3 } \end{array} \right| = 0$$
(x - 1 ) (-3- 0) - ( y - 1 ) (9- 0)+(z+1)(0+2)=0
- 3x + 3 - 9y + 9 + 2z + 2 = 0
3x + 9y - 2z = 14
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