JEE MAIN - Mathematics (2012 - No. 13)
If the line $${{x - 1} \over 2} = {{y + 1} \over 3} = {{z - 1} \over 4}$$ and $${{x - 3} \over 1} = {{y - k} \over 2} = {z \over 1}$$ intersect, then $$k$$ is equal to :
$$-1$$
$${2 \over 9}$$
$${9 \over 2}$$
$$0$$
Explanation
Given lines in vector form are
$$\overrightarrow r = \left( {\widehat i - \overrightarrow j + \overrightarrow k } \right) + \lambda \left( {2\overrightarrow i + 3\overrightarrow j + 4\overrightarrow j } \right)$$
and $$\overrightarrow r = \left( {3\widehat i + k\widehat j} \right) + \mu \left( {\widehat i + 2\widehat j + \widehat k} \right)$$
These will intersect if shortest distance between them $$=0$$
i.e.$$\left( {{{\overrightarrow a }_2} - {{\overrightarrow a }_1}} \right).\overrightarrow {{b_1}} \times {\overrightarrow b _2} = 0$$
$$ \Rightarrow \left| {\matrix{ {3 - 1} & {k + 1} & { - 1} \cr 2 & 3 & 4 \cr 1 & 2 & 1 \cr } } \right| = 0$$
$$ \Rightarrow 2\left( { - 5} \right) - \left( {k + 1} \right)\left( { - 2} \right) - 1\left( 1 \right) = 0$$
$$ \Rightarrow k = 9/2$$
$$\overrightarrow r = \left( {\widehat i - \overrightarrow j + \overrightarrow k } \right) + \lambda \left( {2\overrightarrow i + 3\overrightarrow j + 4\overrightarrow j } \right)$$
and $$\overrightarrow r = \left( {3\widehat i + k\widehat j} \right) + \mu \left( {\widehat i + 2\widehat j + \widehat k} \right)$$
These will intersect if shortest distance between them $$=0$$
i.e.$$\left( {{{\overrightarrow a }_2} - {{\overrightarrow a }_1}} \right).\overrightarrow {{b_1}} \times {\overrightarrow b _2} = 0$$
$$ \Rightarrow \left| {\matrix{ {3 - 1} & {k + 1} & { - 1} \cr 2 & 3 & 4 \cr 1 & 2 & 1 \cr } } \right| = 0$$
$$ \Rightarrow 2\left( { - 5} \right) - \left( {k + 1} \right)\left( { - 2} \right) - 1\left( 1 \right) = 0$$
$$ \Rightarrow k = 9/2$$
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