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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
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Find the angle between the following pair of lines: $ \frac{\large x}{\large 2} = \frac{\large y}{\large 2} = \frac{\large z}{\large 1} \: and\: \frac{\large x-5}{\large 4} = \frac{\large y-2}{\large 1} = \frac{\large z - 3}{\large 8} $

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  • Angle between two lines having d.r as $ b_1(l_1, m_1,n_1) \: and \: b_2 (l_2,m_2,n_2)$ is given by $ cos^{-1}\bigg(\large \frac{\overrightarrow  {b_1}.\overrightarrow {b_2}}{\overrightarrow {|b_1}||\overrightarrow {b_2}|}\bigg)$
Given equation of two lines are
$\large\frac{x}{2}=\frac{y}{2}=\frac{z}{1}$ and $\large\frac{x-5}{4}=\frac{y-2}{1}=\frac{z-3}{8}$
$\Rightarrow\: $ the d.r. of the lines are $\overrightarrow b_1 = (2,2,1) \: \overrightarrow b_2 = (4,1,8)$
$|\overrightarrow b_1|=\sqrt{2^2+2^2+1^2}=\sqrt{9} =3$
$|\overrightarrow b_2|=\sqrt{4^2+1^2+8^2}=\sqrt{81}=9$
$\overrightarrow {b_1}.\overrightarrow {b_2}=(2,2,1).(4,1,8)=8+2+8=18$
We know that the angle between the two lines is given by
$ cos^{-1}\bigg(\large \frac{\overrightarrow {b_1}.\overrightarrow {b_2}}{|\overrightarrow {b_1}||\overrightarrow {b_2}|}\bigg)$
The angle between the lines =$ cos^{-1} \bigg(\large\frac{18}{3\times 9}\bigg)$
$ cos^{-1}\bigg(\large\frac{2}{3}\bigg)$


answered May 1, 2013 by rvidyagovindarajan_1
edited May 1, 2013 by rvidyagovindarajan_1

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