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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
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Find the shortest distance between the lines whose vector equations are $ \overrightarrow r = ( 1 - t ) \hat i + ( t - 2 )\hat j + ( 3 - 2t )\hat k$ and $ \overrightarrow r = ( s + 1 )\hat i + ( 2s - 1 )\hat j - (2s + 1)\hat k $

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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 :$\overrightarrow r=(1-t) \hat i+(t-2) \hat j +(3-2t)\hat k$
On rearranging we get,
$\overrightarrow r=( \hat i-2 \hat j+3 \hat k)+t(-\hat i+ \hat j -2\hat k)$
Hence $\overrightarrow {a_1}=(\hat i-2 \hat j+3\hat k)$ and $\overrightarrow {b_1}=(-\hat i+\hat j-2\hat k)$
Let $L_2$ be the line :$\overrightarrow r=(s+1) \hat i+(2s-1) \hat j +(2s+1)\hat k$
On rearranging we get,
$\overrightarrow r=( \hat i- \hat j+ \hat k)+s(\hat i+ 2\hat j -2\hat k)$
Hence $\overrightarrow {a_2}=(\hat i- \hat j-\hat k)$ and $\overrightarrow {b_2}=(\hat i+2\hat j-2\hat k)$
Let us 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 & -2 \\ 1 & 2 & -2 \end{vmatrix}$
On expanding we get,
$=(-2+4)\hat i-(2+2)\hat j+(-2-1) \hat k $
$=(2\hat i-4\hat j-3 \hat k) $
$|\overrightarrow b_1\times\overrightarrow b_2|=\sqrt{(2)^2+(-4)^2+(-3)^2}= \sqrt{29 }$
$\overrightarrow a_2-\overrightarrow a_1=( \hat i-\hat j-\hat k)-(\hat i+2 \hat j+3\hat k)$
$\qquad\qquad=( \hat j-4 \hat k)$
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| $
On substituting the values we get
$d=\bigg|\large\frac{(2 \hat i-4 \hat j-3 \hat k). (\hat j-4 \hat k)}{ \sqrt {29}}\bigg|$
On simplifying we get,
$d=\bigg|\large\frac{2 \times 0-4 \times 1-3 \times (-4)}{ \sqrt {29}}\bigg|$
$=\large\frac{0-4+12}{\sqrt {29}}$
$=\large\frac{8}{\sqrt {29}}$units
Hence the shortest distance between the line is $=\large\frac{8}{\sqrt {29}}$units
answered Jun 9, 2013 by meena.p
 

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