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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
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Find the shortest distance between the lines $ \overrightarrow r = ( \hat i + 2\hat j + \hat k) + \lambda ( \hat i - \hat j + \hat k )$ and $ \overrightarrow r = 2 \hat i - \hat j - \hat k + \mu (2\hat i + \hat j + 2\hat k )$

This question has appeared in model paper 2012

$\begin{array}{1 1} \large\frac{12}{\sqrt{35}} \\\large\frac{9}{\sqrt {35}} \\\large\frac{4 \sqrt 2}{2} \\ \large\frac{3 \sqrt2}{2} \end{array} $

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  • Shortest distance between two lines is given by $d= \bigg| \large\frac{(\overrightarrow b_1 \times \overrightarrow b_2).(\overrightarrow a_2 - \overrightarrow a_1)}{|\overrightarrow b_1 \times \overrightarrow b_2|} \bigg| $
Let $L_1$ be the line :$\overrightarrow r=(\hat i+2 \hat j +\hat k)+\lambda(\hat i-\hat j+\hat k)$
Hence $\overrightarrow {a_1}=\hat i+2 \hat j+\hat k$
The Direction cosines of the vector $a_1$ is $(1,2,1)$
$\overrightarrow {b_1}=\hat i- \hat j+\hat k$
The Direction cosines of the vector $b_1$ is $(1,-1,1)$
Let $L_2$ be the line :$\overrightarrow r=(2 \hat i-2 \hat j -\hat k)+\mu (2\hat i+\hat j+2\hat k)$
Hence $\overrightarrow {a_2}=2\hat i- \hat j-\hat k$
The Direction cosines of the vector $a_2$ is $(2,-1,-1)$
Hence $\overrightarrow {b_2}=2\hat i+ \hat j+2\hat k$
The Direction cosines of the vector $b_2$ is $(2,1,2)$
We know the Shortest distance between two lines is $d= \bigg| \large\frac{(\overrightarrow b_1 \times \overrightarrow b_2).(\overrightarrow a_2 - \overrightarrow a_1)}{|\overrightarrow b_1 \times \overrightarrow b_2|} \bigg| $
Let us first determine $ \overrightarrow b_1\times \overrightarrow b_2$
$ \overrightarrow b_1\times \overrightarrow b_2= \begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & -1 & 1 \\ 2 & 1 & 2 \end{vmatrix}$
On expanding we get,
$=-3\hat i-0+3 \hat k $
$=-3\hat i+3 \hat k $
$|\overrightarrow b_1\times\overrightarrow b_2|=\sqrt{(-3)^2+3^2}=3 \sqrt{2 }$
$\overrightarrow a_2-\overrightarrow a_1=(2 \hat i-\hat j-\hat k)-(\hat i+2 \hat j+\hat k)$
$\qquad\qquad=(\hat i-3 \hat j-2 \hat k)$
now substituting the respective value in $'d'$ we get,
$d=\bigg|\large\frac{(-3 \hat i+3 \hat k). (\hat i-3 \hat j-2 \hat k)}{3 \sqrt 2}\bigg|$
$=\bigg|\large\frac{-3-6}{3 \sqrt 2}\bigg|$
$=\large\frac{3}{\sqrt 2}$
Multiply and dividing by $\sqrt 2$ we get,
$\large\frac{3 \sqrt 2}{2}$
answered Jun 5, 2013 by meena.p
 

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