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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 = ( \hat i + 2\hat j + 3\hat k) + \lambda ( \hat i - 3\hat j + 2\hat k )$ and $ \overrightarrow r = 4 \hat i + 5\hat j + 6\hat k + \mu (2\hat i + 3\hat j + \hat k )$

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 :$\overrightarrow r=(\hat i+2 \hat j +3\hat k)+\lambda(\hat i-3\hat j+2\hat k)$
Hence $\overrightarrow {a_1}=\hat i+2 \hat j+3\hat k$
$\qquad\quad\overrightarrow {b_1}=\hat i- 3\hat j+2\hat k$
Let $L_2$ be the line :$\overrightarrow r=(4 \hat i+5 \hat j +6\hat k)+\mu (2\hat i+3\hat j+\hat k)$
Hence $\overrightarrow {a_2}=4\hat i+5 \hat j+6\hat k$
$\qquad\overrightarrow {b_2}=2\hat i+ 3\hat j+\hat k$
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 & -3 & 2 \\ 2 & 3 & 1 \end{vmatrix}$
On expanding we get,
$=\hat i(-3-6)-\hat j(1-4)+\hat k(3+6)$
$=-9 \hat i+3 \hat j+ 9 \hat k$
$|\overrightarrow b_1\times\overrightarrow b_2|=\sqrt{9^2+3^2+9^2}$
$\qquad\qquad\quad=\sqrt {171}$
$\qquad\qquad\quad=3 \sqrt {19}$
$\overrightarrow a_2-\overrightarrow a_1=(4 \hat i+5\hat j+6\hat k)-(\hat i+2 \hat j+3\hat k)$
$\qquad\qquad=3 \hat i+3 \hat j+3 \hat k$
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| $
now substituting the respective value in $'d'$ we get,
$d=\bigg|\large\frac{(3 \hat i+3 \hat j+3 \hat k). (-9\hat i+3 \hat j+9 \hat k)}{3 \sqrt 19}\bigg|$
On Simplifying we get,
$=\bigg|\large\frac{-27+9+27}{3 \sqrt {19}}\bigg|$
$=\large\frac{3}{\sqrt {19}}$ units
answered Jun 6, 2013 by meena.p
 

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