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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  3-D Geometry
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The shortest distance between the lines $\large\frac{x-6}{1}=\frac{2-y}{2}=\frac{z-2}{2}$ and $\large\frac{x+4}{3}=\frac{y}{-2}=\frac{1-z}{2}$ is?

$(a)\;9\\(b)\;\large\frac{25}{3} \\ (c)\;\large\frac{16}{3}\\(d)\;4 $

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  • S.D. between the lines $\overrightarrow r=\overrightarrow {a_1}+\lambda\overrightarrow { b_1}$ and $\overrightarrow r=\overrightarrow {a_2}+\mu \overrightarrow{ b_2}$ is $\bigg|\large\frac{(\overrightarrow {b_1} \times \overrightarrow {b_2}).(\overrightarrow {a_1}-\overrightarrow {a_2})}{|\overrightarrow {b_1}\times\overrightarrow{ b_2}|}\bigg|$
Writing the given lines in standard vector form we get
$\overrightarrow r=(6\hat i+2\hat j+2\hat k)+\lambda (\hat i-2\hat j+2\hat k)$....(i)
$\overrightarrow r=(-4\hat i+\hat k)+\lambda (3\hat i-2\hat j-2\hat k)$.........(ii)
Here $\overrightarrow a_1=(6,2,2),\:\:\overrightarrow {a_2}=(-4,0,1),\:\:\overrightarrow {b_1}=(1,-2,2),\:\:\overrightarrow b_2=(3,-2,-2)$
$\overrightarrow a_1-\overrightarrow a_2=(10,2,1)$
$\overrightarrow b_1\times \overrightarrow b_2=\left |\begin {array}{ccc} \hat i & \hat j & \hat k \\1 & -2 & 2\\ 3 & -2 & -2\end {array} \right |=(8,8,4)$
$|\overrightarrow b_1\times \overrightarrow b_2|=\sqrt {144}=12$
$(\overrightarrow a_1-\overrightarrow a_2).(\overrightarrow b_1\times\overrightarrow b_2)=(10,2,1).(8,8,4)$
S.D.= $\bigg|\large\frac{(\overrightarrow {b_1} \times \overrightarrow {b_2}).(\overrightarrow {a_1}-\overrightarrow {a_2})}{|\overrightarrow {b_1}\times\overrightarrow{ b_2}|}\bigg|$
answered Dec 29, 2013 by rvidyagovindarajan_1

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