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The planes $\overrightarrow{r} .(\overrightarrow{2i}+\lambda\overrightarrow{j}-\overrightarrow{3k})=10 $and $ \overrightarrow{r}. (\lambda\overrightarrow{i}+\overrightarrow{3j}+\overrightarrow{k})=5$ are perpendiculare find $\lambda$

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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+\lambda\overrightarrow j-3\overrightarrow k)=10, \: \overrightarrow r.(\lambda\overrightarrow i+3\overrightarrow j+\overrightarrow k)=5$
The normal vectors are $ \overrightarrow n_1 = 2\overrightarrow i+\lambda\overrightarrow j-3\overrightarrow k,\: \overrightarrow n_2=\lambda\overrightarrow i+3\overrightarrow j+\overrightarrow k$
planes are at right angles
$ \therefore \overrightarrow n_1.\overrightarrow n_2=0 \Rightarrow (2)(\lambda)+(\lambda)(3)+(-3)(1)=0$
$ \therefore 5\lambda-3=0 \Rightarrow \lambda = \large\frac{3}{5}$
answered Jun 17, 2013 by thanvigandhi_1

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