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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Find the shortest distance between the following pairs of lines whose vector equation are $ \overrightarrow r = 3\hat i + 8\hat j + 3\hat k + \lambda(3\hat i - \hat j + \hat k)\: and \: \overrightarrow r = -3\hat i - 7\hat j + 6\hat k + \mu(-3\hat i + 2\hat j + 4\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| $
Step 1:
$\overrightarrow{r}=3\hat i+8\hat j+3\hat k+\lambda(3\hat i-\hat j+\hat k)$
$\overrightarrow{r}=-3\hat i-7\hat j+6\hat k+\mu(-3\hat i+2\hat j+4\hat k)$
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| $
Here $\overrightarrow {a_1}=3\hat i+8\hat j+3\hat k$
$\overrightarrow {a_2}=-3\hat i-7\hat j+6\hat k$
$\overrightarrow {b_1}=3\hat i-\hat j+4\hat k$
$\overrightarrow {b_2}=-3\hat i+2\hat j+4\hat k$
Step 2:
$\overrightarrow {b_1}\times \overrightarrow {b_2}=\begin{vmatrix}\hat i &\hat j&\hat k\\3 &-1 &1\\-3 & 2& 4\end{vmatrix}$
$\qquad\quad=\hat i(-4-2)-\hat j(12+3)+\hat k(6-3)$
$\qquad\quad=-6\hat i-15\hat j+3\hat k$
$\mid \overrightarrow {b_1}\times \overrightarrow {b_2}\mid=\sqrt{(-6)^2+(-15)^2+(3)^2}$
$\qquad\qquad=\sqrt{270}$
Step 3:
$\overrightarrow {a_2}-\overrightarrow {a_1}=(-3\hat i-7\hat j+6\hat k)-(3\hat i+8\hat j+3\hat k)$
$\qquad\quad=-6\hat i-15\hat j+3\hat k$
Step 4:
$d=\large\frac{(-6\hat i-15\hat j+3\hat k).(-6\hat i-15\hat j+3\hat k)}{\sqrt{270}}$
$\;\;=\large\frac{36+225+9}{\sqrt{270}}$
$\;\;=\large\frac{270}{\sqrt{270}}$
$\;\;=\sqrt{270}$units
$\;\;=3\sqrt{30}$units
answered Sep 30, 2013 by sreemathi.v
 

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