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.

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

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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