JEE MAIN - Mathematics (2008 - No. 26)
If the straight lines $$\,\,\,\,\,$$ $$\,\,\,\,\,$$ $${{x - 1} \over k} = {{y - 2} \over 2} = {{z - 3} \over 3}$$ $$\,\,\,\,\,$$ and$$\,\,\,\,\,$$ $${{x - 2} \over 3} = {{y - 3} \over k} = {{z - 1} \over 2}$$ intersects at a point, then the integer $$k$$ is equal to
$$-5$$
$$5$$
$$2$$
$$-2$$
Explanation
The two lines intersect if shortest distance between them is zero $$i.e.$$
$${{\left( {{{\overrightarrow a }_2} - {{\overrightarrow a }_1}} \right).{{\overrightarrow b }_1} \times {{\overrightarrow b }_2}} \over {\left| {{{\overrightarrow b }_1} \times {{\overrightarrow b }_2}} \right|}} = 0$$
$$ \Rightarrow \left( {{{\overrightarrow a }_2} - {{\overrightarrow a }_1}} \right).{\overrightarrow b _1} \times {\overrightarrow b _2} = 0$$
where $${\overrightarrow a _1} = \widehat i + 2\widehat j + 3\widehat k,{\overrightarrow b _1} = k\widehat i + 2\widehat j + 3\widehat k$$
$$\,\,\,\,\,\,\,{\overrightarrow a _2} = 2\widehat i + 3\widehat j + \widehat k,\,\,{\widehat b_2} = 3\widehat i + k\widehat j + 2\widehat k$$
$$ \Rightarrow \left| {\matrix{ 1 & 1 & { - 2} \cr k & 2 & 3 \cr 3 & k & 2 \cr } } \right| = 0$$
$$ \Rightarrow 1\left( {4 - 3k} \right) - 1\left( {2k - 9} \right) - 2\left( {{k^2} - 6} \right) = 0$$
$$ \Rightarrow - 2{k^2} - 5k + 25 = 0 \Rightarrow k = - 5$$ $$\,\,\,\,$$ or $$\,\,\,\,$$ $${5 \over 2}$$
As $$k$$ is an integer, therefore $$k=-5$$
$${{\left( {{{\overrightarrow a }_2} - {{\overrightarrow a }_1}} \right).{{\overrightarrow b }_1} \times {{\overrightarrow b }_2}} \over {\left| {{{\overrightarrow b }_1} \times {{\overrightarrow b }_2}} \right|}} = 0$$
$$ \Rightarrow \left( {{{\overrightarrow a }_2} - {{\overrightarrow a }_1}} \right).{\overrightarrow b _1} \times {\overrightarrow b _2} = 0$$
where $${\overrightarrow a _1} = \widehat i + 2\widehat j + 3\widehat k,{\overrightarrow b _1} = k\widehat i + 2\widehat j + 3\widehat k$$
$$\,\,\,\,\,\,\,{\overrightarrow a _2} = 2\widehat i + 3\widehat j + \widehat k,\,\,{\widehat b_2} = 3\widehat i + k\widehat j + 2\widehat k$$
$$ \Rightarrow \left| {\matrix{ 1 & 1 & { - 2} \cr k & 2 & 3 \cr 3 & k & 2 \cr } } \right| = 0$$
$$ \Rightarrow 1\left( {4 - 3k} \right) - 1\left( {2k - 9} \right) - 2\left( {{k^2} - 6} \right) = 0$$
$$ \Rightarrow - 2{k^2} - 5k + 25 = 0 \Rightarrow k = - 5$$ $$\,\,\,\,$$ or $$\,\,\,\,$$ $${5 \over 2}$$
As $$k$$ is an integer, therefore $$k=-5$$
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