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Find the angle between following planes; $2x-3y+4z=1$ and $-x+y=4$

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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-3y+4z=1$ (i) and $-x+y=4$ (ii)
$ \overrightarrow n_1 = 2\overrightarrow i-3\overrightarrow j+4\overrightarrow k \: \: \overrightarrow n_2=-\overrightarrow i+\overrightarrow j$
$ \theta = \cos^{-1} \large\frac{[\overrightarrow n_1.\overrightarrow n_2]}{|\overrightarrow n_1||\overrightarrow n_2|}=\cos^{-1} \bigg[ \large\frac{(2)(-1)+(-3)(1)+0}{\sqrt{4+9+16}\sqrt{1+1}} \bigg]$
$ = \cos^{-1} \large\frac{-5}{\sqrt{58}}$
answered Jun 17, 2013 by thanvigandhi_1
 

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