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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
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Find the angle between the pairs of lines: $ \overrightarrow r = 3\hat i + \hat j - 2\hat k + \lambda (\hat i - \hat j - 2\hat k)\: and \: \overrightarrow r = 2\hat i - \hat j - 56\hat k + \mu (3\hat i - 5\hat j - 4\hat k) $

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  • Angle betwen two lines $\overrightarrow r=\overrightarrow {a_1}+\lambda\overrightarrow {b_1}$ and $\overrightarrow r=\overrightarrow {a_2}+\mu\overrightarrow {b_2}$ is given by given by $ cos^{-1}\bigg( \frac{\overrightarrow {b_1}.\overrightarrow {b_2}}{|\overrightarrow {b_1}||\overrightarrow {b_2}|}\bigg)$.
 
Let given two lines be
$\overrightarrow r=(3\hat i+\hat j-2\hat k)+\lambda(\hat i-\hat j-2\hat k)$ .........$(L_1)$
$\overrightarrow r=(2\hat i-\hat j+56\hat k)+\mu(3\hat i-5\hat j-4\hat k)$ .........$(L_2)$
$\Rightarrow\:d.r.\: of \: L_1 = \overrightarrow b_1 = (1,-1,-2)$
d.r of $L_2 = \overrightarrow b_2 = (3,-5,-4)$
$|\overrightarrow {b_1}|=\sqrt{1^2+(-1)^2+(-2)^2}=\sqrt {6}$
$|\overrightarrow {b_2}|=\sqrt{3^2+(-5)^2+(-4)^2}=\sqrt {50}=$
$\overrightarrow {b_1}.\overrightarrow {b_2}=(1,-1,-2).(3,-5,-4)=3+5+8=16$
We know that Angle betwen two lines $\overrightarrow r=\overrightarrow {a_1}+\lambda\overrightarrow {b_1}$ and $\overrightarrow r=\overrightarrow {a_2}+\mu\overrightarrow {b_2}$ is given by given by $ cos^{-1}\bigg( \frac{\overrightarrow {b_1}.\overrightarrow {b_2}}{|\overrightarrow {b_1}||\overrightarrow {b_2}|}\bigg)$.
$\Rightarrow$ Angle between $L_1\:and\:L_2$ is given by
$ cos^{-1}\large\bigg(\frac{16}{\sqrt 6\sqrt{50}}\bigg)=cos^{-1}\bigg(\frac{8}{5\sqrt 3}\bigg)$

 

answered Apr 30, 2013 by rvidyagovindarajan_1
edited Apr 30, 2013 by rvidyagovindarajan_1
 

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