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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Find the distance between the two lines $l_1$ and $l_2$ where $l_1: \overrightarrow r=(\hat i+2\hat j-4\hat k)+\lambda(2\hat i+3\hat j+6\hat k)$ and $l_2: \overrightarrow r=(3\hat i+3\hat j-5\hat k)+\mu(4\hat i+6\hat j+12\hat k)$

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  • General vector equation of a line is $\overrightarrow r=\overrightarrow a+\lambda\overrightarrow b$
  • Distance between two parallel lines $\overrightarrow r=\overrightarrow {a_1}+\lambda\overrightarrow {b_1}$ and $\overrightarrow r=\mu\overrightarrow {b_1}$ is given by $\bigg|\large\frac{\overrightarrow b_1\times(\overrightarrow a_2-\overrightarrow a_1)}{|\overrightarrow b_1|}\bigg|$
Given lines are
$l_1: \overrightarrow r=(\hat i+2\hat j-4\hat k)+\lambda(2\hat i+3\hat j+6\hat k)$ and
$l_2: \overrightarrow r=(3\hat i+3\hat j-5\hat k)+\mu(4\hat i+6\hat j+12\hat k)$
Comparing the equations with the general vector equation of a line we get
$\overrightarrow {a_1}=\hat i+2\hat j-4\hat k$
$\overrightarrow {a_2}=3\hat i+3\hat j-5\hat k$
$\overrightarrow {b_1}=2\hat i+3\hat j+6\hat k$
$\overrightarrow {b_2}=4\hat i+6\hat j+12\hat k=2.\overrightarrow {b_1}$
$\therefore \overrightarrow {b_1}$ is $||$ to $\overrightarrow {b_2}$
$i.e.,$ $l_1$ is parallel to $l_2$
Distance between two parallel lines $\overrightarrow r=\overrightarrow {a_1}+\lambda\overrightarrow {b_1}$ and $\overrightarrow r=\mu\overrightarrow {b_1}$ is given by $\bigg|\large\frac{\overrightarrow b_1\times(\overrightarrow a_2-\overrightarrow a_1)}{|\overrightarrow b_1|}\bigg|$
$\overrightarrow a_2-\overrightarrow a_1=(3\hat i+3\hat j-5\hat k)-(\hat i+2\hat j-4\hat k)$
$=2\hat i+\hat j-\hat k$
$\overrightarrow b_1\times (\overrightarrow a_2-\overrightarrow a_1)=\left|\begin {array}{ccc}\hat i &\hat j & \hat k\\2 &3 & 6\\2 &1 &-1\end {array}\right|$
$=(-3-6)\hat i-(-2-12)\hat j+(2-6)\hat k$
$=-9\hat i+14\hat j-4\hat k$
$\Rightarrow\:|\overrightarrow b_1\times(\overrightarrow a_2-\overrightarrow a_1)|=\sqrt {(-9)^2+14^2+4^2}=\sqrt {289}=17$
$|\overrightarrow b_1|=\sqrt {2^2+3^2+6^2}=\sqrt {49}=7$
$\therefore\:\bigg|\large\frac{\overrightarrow b_1\times(\overrightarrow a_2-\overrightarrow a_1)}{|\overrightarrow b_1|}\bigg|$$=\large\frac{17}{7}$
$\Rightarrow\:$ Distance between $l_1$ and $ l_2$ is $ \large\frac{17}{7}$
answered Mar 21, 2014 by rvidyagovindarajan_1
 

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