Question

The plane 2x – 3y + 6z – 11 = 0 makes an angle $ {\sin ^{ - 1}}\left( \alpha \right) $ with the X – axis. The value of $ \alpha $ is equal to
A. $ \dfrac{2}{7} $
B. $ \dfrac{3}{7} $
C. \[\dfrac{{\sqrt 2 }}{7}\]
D. \[\dfrac{{\sqrt 3 }}{7}\]

Answer 1

Hint: Before attempting this question, one should have prior knowledge about the concept line and plane also remember to use \[\sin \theta = \dfrac{{\left| {\overrightarrow a .\overrightarrow n } \right|}}{{\left| {\overrightarrow a } \right|\left| {\overrightarrow n } \right|}}\]where $ \theta $ is the angle between line and plane, use this information to approach the solution.

Complete step-by-step answer:
The given equation of the plane is 2x - 3y + 6z - 11 = 0
We know that the normal vector of the plane can be written as $ \overrightarrow n = 2\widehat i - 3\widehat j + 6\widehat k $
Since the angle is made along the X - axis, therefore, y = 0, z = 0
So, the line vector comes out to be $ \overrightarrow a = \widehat i + 0\widehat j + 0\widehat k $
The angle between the line and plane is \[\sin \theta = \dfrac{{\left| {\overrightarrow a .\overrightarrow n } \right|}}{{\left| {\overrightarrow a } \right|\left| {\overrightarrow n } \right|}}\]
Substituting the values in the above formula we get
\[\sin \theta = \dfrac{{\left| {\left( {\widehat i + 0\widehat j + 0\widehat k} \right).\left( {2\widehat i - 3\widehat j + 6\widehat k} \right)} \right|}}{{\left| {\left( {\widehat i + 0\widehat j + 0\widehat k} \right)} \right|\left| {\left( {2\widehat i - 3\widehat j + 6\widehat k} \right)} \right|}}\]
Also we know that dot product of any two vector $ \overrightarrow p = {x_1}\widehat i + {y_1}\widehat j + {z_1}\widehat k $ and $ \overrightarrow q = {x_2}\widehat i + {y_2}\widehat j + {z_2}\widehat k $ is given as; $ \overrightarrow p .\overrightarrow q = \left( {{x_1} \times {x_2}} \right)\widehat i + \left( {{y_1} \times {y_2}} \right)\widehat j + \left( {{z_1} \times {z_2}} \right)\widehat k $
And magnitude of any vector $ \overrightarrow p = x\widehat i + y\widehat j + z\widehat k $ is given as; $ \left| {\overrightarrow p } \right| = \sqrt {{x^2} + {y^2} + {z^2}} $
So, \[\sin \theta = \dfrac{{\left| {\widehat i \times 2\widehat i + 0\widehat j \times \left( { - 3\widehat j} \right) + 0\widehat k \times 6\widehat k} \right|}}{{\sqrt 1 \sqrt {{{\left( 2 \right)}^2} + {{\left( 3 \right)}^2} + {{\left( 6 \right)}^2}} }}\]
 $ \Rightarrow $ \[\sin \theta = \dfrac{{\left| {2\widehat i} \right|}}{{\sqrt 1 \sqrt {{{\left( 2 \right)}^2} + {{\left( 3 \right)}^2} + {{\left( 6 \right)}^2}} }}\]
 $ \Rightarrow $ \[\sin \theta = \dfrac{{\sqrt {{{\left( 2 \right)}^2}} }}{7} = \dfrac{2}{7}\]or \[\theta = {\sin ^{ - 1}}\left( {\dfrac{2}{7}} \right)\]
Since plane makes an angle of $ {\sin ^{ - 1}}\left( \alpha \right) $ with X-axis
Therefore, $ \theta = {\sin ^{ - 1}}\left( \alpha \right) $
Now substituting the value in the above equation, we get
 $ {\sin ^{ - 1}}\left( {\dfrac{2}{7}} \right) = {\sin ^{ - 1}}\left( \alpha \right) $
 $ \Rightarrow $ $ \alpha = \dfrac{2}{7} $
Therefore, the value of $ \alpha $ is equal to $ \dfrac{2}{7} $

So, the correct answer is “Option A”.

Note: In the above solution we used the basic concept of vector, always keep in mind that if a plane makes an angle with the X-axis which means the Y and Z coordinates of that line will be zero. So, as we got the equation of line at X-axis then we apply the formula of angle between plane and a line and directly apply the properties of vector like dot product of two vectors and magnitude of vector PQ is represented by $ \left| {PQ} \right| $ .