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The plane 2x - 3y + 6z - 11 = 0 makes an angle \sin^{-1}(\alpha)  with the x-axis. The value of α is:
A. \frac{\sqrt{3}}{2}
B. \frac{\sqrt{2}}{3}
C. \frac{2}{7}
D. \frac{3}{7}

Answers (1)

Given, the equation of the plane is 2x - 3y + 6z - 11 = 0.

The normal to this plane is,


Also, the x-axis has the direction vector \vec{b}=\hat{i}

Also, we know that the angle \varphi  between the line with direction vector b and the plane having the normal vector n is:

\\ \sin \varphi =\left | \frac{\vec{b}.\vec{n}}{\left |\vec{b} \right |.\left |\vec{n} \right |} \right |\\ \Rightarrow \sin \varphi=\left | \frac{1(2)+0(-3)+0(6)}{\sqrt{2^{2}+3^{2}+6^{2}}\sqrt{1^{2}+0^{2}+0^{2}}} \right |\\ \Rightarrow \sin \varphi = \left | \frac{2}{\sqrt{49}} \right |=\frac{2}{7}\\\Rightarrow \varphi =\sin^{-1}\left ( \frac{2}{7} \right )
On comparing, we find\alpha =\frac{2}{7}  (Option C)

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