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Home  >>  TN XII Math  >>  Vector Algebra
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Find the angle between the followin planes: $2x+y-z=9 $and $x+2y+z=7$

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  • The angle between two planes is the angle between the normals to the two planes. $ \theta=\cos^{-1}\bigg( \large\frac{\overrightarrow n_1.\overrightarrow n_2}{|\overrightarrow n_1||\overrightarrow n_2|} \bigg).$ If $ \overrightarrow n_1.\overrightarrow n_2=0$ the planes are at right angles.
$ 2x+y-z=9$ (i) $x+2y+z=7$ (ii)
The normal vectors to the planes are
$ \overrightarrow n_1=2\overrightarrow i+\overrightarrow j-\overrightarrow k\: and \: \overrightarrow n_2=\overrightarrow i+2\overrightarrow j+\overrightarrow k$
$ \therefore \theta = \cos^{-1} \bigg[ \large\frac{\overrightarrow n_1.\overrightarrow n_2}{|\overrightarrow n_1||\overrightarrow n_2|} \bigg] = \cos^{-1} \bigg[ \large\frac{(2)(1)+(1)(2)+(-1)(1)}{\sqrt{4+1+1}\sqrt{1+4+1}} \bigg]$
$ = \cos^{-1} \large\frac{3}{6}=\cos^{-1}\large\frac{1}{2}=\large\frac{\pi}{3}$
answered Jun 17, 2013 by thanvigandhi_1
 

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