Browse Questions

# The plane 2x-3y+6z-11=0 makes an angle $\sin^{-1}(\alpha)$ with x-axis.The value of $\alpha$ is equal to

$(A)\;\frac{3}{2}\qquad(B)\;\frac{\sqrt 2}{3}\qquad(C)\;\frac{2}{7}\qquad(D)\;\frac{3}{7}$

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.