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Find the angle between following plane:$r=(3i+j-k)=7 $ and $r=(i+4j-2k)=10$

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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.
$ \overrightarrow r .(3\overrightarrow i+\overrightarrow j-\overrightarrow k)=15\: and \: \overrightarrow r.(\overrightarrow i+4\overrightarrow j-2\overrightarrow k)=10$
$ \overrightarrow n_1=3\overrightarrow i+\overrightarrow j-\overrightarrow k, \: \: \overrightarrow n_2=\overrightarrow i+4\overrightarrow j-2\overrightarrow k$ are the normal vectors.
$ \theta = \cos^{-1} \bigg[ \large\frac{\overrightarrow n_1.\overrightarrow n_2}{|\overrightarrow n_1||\overrightarrow n_2|} \bigg] = \cos^{-1} \bigg[ \large\frac{(3)(1)+(1)(4)+(-1)(-2)}{\sqrt{9+1+1}\sqrt{1+16+4}} \bigg]$
$ = \cos^{-1} \large\frac{3+4+2}{\sqrt{231}}=\cos^{-1} \large\frac{9}{\sqrt{231}}$
answered Jun 17, 2013 by thanvigandhi_1
 

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