JEE MAIN - Mathematics (2004 - No. 49)
If the straight lines
$$x=1+s,y=-3$$$$ - \lambda s,$$ $$z = 1 + \lambda s$$ and $$x = {t \over 2},y = 1 + t,z = 2 - t,$$ with parameters $$s$$ and $$t$$ respectively, are co-planar, then $$\lambda $$ equals :
$$x=1+s,y=-3$$$$ - \lambda s,$$ $$z = 1 + \lambda s$$ and $$x = {t \over 2},y = 1 + t,z = 2 - t,$$ with parameters $$s$$ and $$t$$ respectively, are co-planar, then $$\lambda $$ equals :
$$0$$
$$-1$$
$$ - {1 \over 2}$$
$$-2$$
Explanation
The given lines are
$$x - 1 = {{y + 3} \over { - \lambda }} = {{z - 1} \over \lambda } = s.........\left( 1 \right)$$
and $$2x = y - 1 = {{z - 2} \over { - 1}} = t..........\left( 2 \right)$$
The lines are coplanar, if
$$\left| {\matrix{ {0 - \left( { - 1} \right)} & { - 1 - 3} & { - 2 - \left( { - 1} \right)} \cr 1 & { - \lambda } & \lambda \cr {{1 \over 2}} & 1 & { - 1} \cr } } \right| = 0$$
$${c_2} \to {c_2} + {c_3};\left| {\matrix{ 1 & { - 5} & 1 \cr 1 & 0 & \lambda \cr {{1 \over 2}} & 0 & { - 1} \cr } } \right| = 0$$
$$ \Rightarrow 5\left( { - 1 - {\lambda \over 2}} \right) = 0 \Rightarrow \lambda = - 2$$
$$x - 1 = {{y + 3} \over { - \lambda }} = {{z - 1} \over \lambda } = s.........\left( 1 \right)$$
and $$2x = y - 1 = {{z - 2} \over { - 1}} = t..........\left( 2 \right)$$
The lines are coplanar, if
$$\left| {\matrix{ {0 - \left( { - 1} \right)} & { - 1 - 3} & { - 2 - \left( { - 1} \right)} \cr 1 & { - \lambda } & \lambda \cr {{1 \over 2}} & 1 & { - 1} \cr } } \right| = 0$$
$${c_2} \to {c_2} + {c_3};\left| {\matrix{ 1 & { - 5} & 1 \cr 1 & 0 & \lambda \cr {{1 \over 2}} & 0 & { - 1} \cr } } \right| = 0$$
$$ \Rightarrow 5\left( { - 1 - {\lambda \over 2}} \right) = 0 \Rightarrow \lambda = - 2$$
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