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Home  >>  TN XII Math  >>  Vector Algebra
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Show that the following planes are at right angles.$\overrightarrow{r}(\overrightarrow{2i}-\overrightarrow{j}+\overrightarrow{k})=15$ and$ \overrightarrow{r}(\overrightarrow{i}-\overrightarrow{j}-\overrightarrow{3k})=3$

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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 .(2\overrightarrow i-\overrightarrow j+\overrightarrow k)=15\: and \: \overrightarrow r.(\overrightarrow i-\overrightarrow j-3\overrightarrow k)=3$
The normal vectors to the planes are
$ \overrightarrow n_1 = 2\overrightarrow i-\overrightarrow j+\overrightarrow k\: \: \overrightarrow n_2 = \overrightarrow i-\overrightarrow j-3\overrightarrow k$
Consider $ \overrightarrow n_1.\overrightarrow n_2 = (2)(1)+(-1)(-1)+1(-3)$
$ = 2+1-3=0$
$\overrightarrow n_1.\overrightarrow n_2=0 \Rightarrow$ the planes are at right angles.
answered Jun 17, 2013 by thanvigandhi_1
 

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