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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
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Find the shortest distance between the lines given by $\overrightarrow{r}=(8+3\lambda)\hat i-(9+16\lambda)\hat j+(10+7\lambda)\hat k\; and\; \overrightarrow{r}=15\hat i+29\hat j+5\hat k+\mu(3\hat i+8\hat j-5\hat k)$

$\begin{array}{1 1}14 \\ 10 \\ 28 \\ 7 \end{array} $

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  • Shortest distance between the line is $ d = \bigg| \large\frac{(\overrightarrow a_2-\overrightarrow a_1).(\overrightarrow b_1 \times \overrightarrow b_2)}{| \overrightarrow b_1 \times \overrightarrow b_2 |} \bigg|$
Step 1
The given lines are:
$L_1: \overrightarrow r = (8+3 \lambda) \hat i - (9+16 \lambda ) \hat j + (10 +7 \lambda)\hat k$
$L_2: \overrightarrow r = 15 \hat i +29 \hat j + 5 \hat k+\mu (3 \hat i+8 \hat j-5 \hat k)$
Line $L_1$ can be written as
$ \overrightarrow r = 8 \hat i - 9 \hat j + 10 \hat k+\lambda(3 \hat i-16 \hat j+7 \hat k)$
We know the Shortest distance between the line is
$ d = \bigg| \large\frac{(\overrightarrow a_2-\overrightarrow a_1).(\overrightarrow b_1 \times \overrightarrow b_2)}{| \overrightarrow b_1 \times \overrightarrow b_2 |} \bigg|$
Here $a_1=8 \hat i-9 \hat j+10 \hat k$
$\qquad a_2=15 \hat i+29 \hat j+5 \hat k$
$\qquad b_1=3 \hat i-16 \hat j+7 \hat k$
$\qquad b_2=3 \hat i+8 \hat j-5 \hat k$
Let us obtain $(a_2-a_1)$
$ (\overrightarrow a_2 - \overrightarrow a_1) = (15 \hat i+29 \hat j+ 5 \hat k)-(8 \hat i-9 \hat j+10 \hat k)$
$\qquad\qquad\quad=7 \hat i+38 \hat j-5 \hat k$
Step 2:
Next let us obtain $ \overrightarrow b_1$ x $ \overrightarrow b_2 $
$ \overrightarrow b_1$ x $ \overrightarrow b_2 =\begin {vmatrix} \hat i & \hat j & \hat k \\ 3 & -16 & 7 \\ 3 & 8 & -5 \end {vmatrix}$
$\qquad\qquad=\hat i(80-56)-\hat j(-15-21)+\hat k(24+48)$
$\qquad\qquad=24 \hat i+36 \hat j+72 \hat k$
$\qquad\qquad=12(2\hat i+3 \hat j+6\hat k)$
$| \overrightarrow b_1$ x $ \overrightarrow b_2 |=\sqrt {2^2+3^2+6^2}$
$\qquad\qquad\quad=\sqrt {4+9+36}$
$\qquad\qquad\quad=7$
Step 3:
Now substituting the respective values we get,
$d=\Bigg|\large\frac{(7 \hat i+38 \hat j-5 \hat k).(24 \hat i+36 \hat j+72 \hat k)}{84}\Bigg|$
$\quad=\Bigg|\large\frac{168+1368-360}{84}\Bigg|$
$\quad =\large\frac{1176}{84}$$=14$
Hence the shortest distance between the line is $14$
answered Jun 11, 2013 by meena.p
 

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