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Find the angle between the line $\large\frac{x-2}{3}=\frac{y+1}{-1}=\frac{z-3}{-2}$and the plane $3x+4y+z+5=0$

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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 $ \large\frac{x-2}{3}=\large\frac{y+1}{-1}=\large\frac{z-3}{-2}$ is parallel to the vector $ \overrightarrow u=3\overrightarrow i-\overrightarrow j-2\overrightarrow k$.
The normal vector to the plane
$3x+4y+z+5=0 \: is \: \overrightarrow n=3\overrightarrow i+4\overrightarrow j+\overrightarrow k$
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{(3)(3)+(-1)(4)+(-2)(1)}{\sqrt{9+1+4}\sqrt{9+16+1}} \bigg]$
$ = \sin^{-1} \bigg[ \large\frac{9-4-2}{\sqrt{14}\sqrt{26}} \bigg]$
$ = \sin^{-1} \large\frac{3}{2\sqrt{91}}$
answered Jun 18, 2013 by thanvigandhi_1

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