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Find the angle between the line $\overrightarrow{r}=\overrightarrow{i}+\overrightarrow{j}+\overrightarrow{3k}+\lambda(\overrightarrow{2i}+\overrightarrow{j}-\overrightarrow{k})$ and the plane $\overrightarrow{r}.(\overrightarrow{i}+\overrightarrow{j})=1.$

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  • The angle between a line and a plane is the complement of the angle between the line and the normal to the plane $ \theta = \sin^{-1} \bigg(\large\frac{\overrightarrow b.\overrightarrow n}{|\overrightarrow b||\overrightarrow n|} \bigg)$ where the line is parallel to $ \overrightarrow b\: and \: \overrightarrow n$ is the normal vector to the plane.
The line $ \overrightarrow r= \overrightarrow i+\overrightarrow j+3\overrightarrow k+ \lambda(2\overrightarrow i+\overrightarrow j-\overrightarrow k)$ is parallel to $ \overrightarrow u=2\overrightarrow i+\overrightarrow j-\overrightarrow k$
and the normal vector to the plane $ \overrightarrow r.(\overrightarrow i+\overrightarrow j)=1$ is $ \overrightarrow n=\overrightarrow i+\overrightarrow j$
the angle between the line and the plane is $ \theta = \sin^{-1} \bigg[ \large\frac{\overrightarrow u.\overrightarrow n}{|\overrightarrow u||\overrightarrow n|} \bigg] = \sin^{-1} \bigg[ \large\frac{(2)(1)+(1)(1)}{\sqrt{4+1+1}\sqrt 2} \bigg]$
$ = \sin^{-1} \large\frac{3}{\sqrt{12}} = \sin^{-1} \large\frac{\sqrt 3}{2}$
$ = \large\frac{\pi}{3}$
answered Jun 18, 2013 by thanvigandhi_1

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