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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
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The plane 2x-3y+6z-11=0 makes an angle $\sin^{-1}(\alpha)$ with x-axis.The value of $\alpha$ is equal to

\[(A)\;\frac{3}{2}\qquad(B)\;\frac{\sqrt 2}{3}\qquad(C)\;\frac{2}{7}\qquad(D)\;\frac{3}{7}\]
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Toolbox:
  • Angle between a plane and a line is $ sin\theta = \frac{\overrightarrow b.\overrightarrow n}{|\overrightarrow b||\overrightarrow n|}$
  • Where $\overrightarrow n$ is the normal vector to the plane
Given plane is $ 2x-3y+6z-11=0$
We know $ sin\theta = \frac{\overrightarrow b.\overrightarrow n}{|\overrightarrow b||\overrightarrow n|}$
Here $\overrightarrow n$ is the normal vector to the plane
$\overrightarrow n= 2\hat i-3 \hat j+6 \hat k$
and $\overrightarrow b$ is along x axis.
$(ie) \hat i $ therefore $\sin \theta=\large\frac{(2 \hat i-3 \hat j+6 \hat k).\hat i}{\sqrt {2^2+(-3)^2+6^2}.\sqrt {12}}$
$=\large\frac{2}{\sqrt{49}}=\frac{2}{7}$
Hence $C$ is the correct option.
answered Jun 12, 2013 by meena.p
 
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