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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
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Find the shortest distance between the lines: $ \frac{ \large x + 1 }{ \large 7} = \frac{ \large y + 1}{ \large -6} = \frac{ \large z + 1}{ \large 1}$ and $ \frac{ \large x - 3 }{ \large 1} = \frac{ \large y - 5}{ \large -2} = \frac{ \large z - 7}{ \large 1}$

This question has appeared in model paper 2012.

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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 :$\large\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}$
Here Direction cosines of the vector $a_1$ is $(-1,-1,-1)$
and Direction cosines of the vector $b_1$ is $(7,-6,1)$
Let $L_2$ be the line :$\large\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}$
Here Direction cosines of the vector $a_2$ is $(3,5,7)$
and direction cosines of the vector $b_2$ is $(1,-2,1)$
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 \\ 7 & -6 & 1 \\ 1 & -2 & 1 \end{vmatrix}$
On expanding we get,
$=\hat i(-6 \times 1 -1 \times -2)-\hat j(7 \times 1-1 \times 1)+ \hat k(7 \times -2 -1 \times -6)$
On simplifying we get,
$=4\hat i-6 \hat j- \hat k $
$|\overrightarrow b_1\times\overrightarrow b_2|=\sqrt{(-4)^2+(-6)^2+(-8)^2}$
$\qquad\qquad\quad= \sqrt{16+36+64 }$
$\qquad\qquad\quad= \sqrt{116 }$
$\qquad\qquad\quad= 2\sqrt{29 }$
$\overrightarrow a_2-\overrightarrow a_1=(3+1) \hat i+(5+1)\hat j+(7+1)\hat k$
$\qquad\qquad=4 \hat i+6\hat j+8\hat k$
now substituting the respective value in $'d'$ we get,
$d=\bigg|\large\frac{(-4 \hat i-6 \hat j-\hat k). (4 \hat i+6 \hat j+8 \hat k)}{ \sqrt {116}}\bigg|$
$\quad=\bigg|\large\frac{-16-36-64}{2 \sqrt {29}}\bigg|$
$\quad=\bigg|\large\frac{116}{2 \sqrt 29}\bigg|$
Hence the shortest distance between the lines is=$2 \sqrt {29}$
answered Jun 9, 2013 by meena.p
 

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