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Question: The plane 2x – 3y + 6z – 11 = 0 makes an angle \( {\sin ^{ - 1}}\left( \alpha \right) \) with the X ...

The plane 2x – 3y + 6z – 11 = 0 makes an angle sin⁡−1(α){\sin ^{ - 1}}\left( \alpha \right) with the X – axis. The value of α\alpha is equal to
A. 27\dfrac{2}{7}
B. 37\dfrac{3}{7}
C. 27\dfrac{{\sqrt 2 }}{7}
D. 37\dfrac{{\sqrt 3 }}{7}

Explanation

Solution

Before attempting this question, one should have prior knowledge about the concept line and plane also remember to use sin⁡θ=∣a→.n→∣∣a→∣∣n→∣\sin \theta = \dfrac{{\left| {\overrightarrow a .\overrightarrow n } \right|}}{{\left| {\overrightarrow a } \right|\left| {\overrightarrow n } \right|}}where θ\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 n→=2i^−3j^+6k^\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 a→=i^+0j^+0k^\overrightarrow a = \widehat i + 0\widehat j + 0\widehat k
The angle between the line and plane is sin⁡θ=∣a→.n→∣∣a→∣∣n→∣\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⁡θ=∣(i^+0j^+0k^).(2i^−3j^+6k^)∣∣(i^+0j^+0k^)∣∣(2i^−3j^+6k^)∣\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 p→=x1i^+y1j^+z1k^\overrightarrow p = {x_1}\widehat i + {y_1}\widehat j + {z_1}\widehat k and q→=x2i^+y2j^+z2k^\overrightarrow q = {x_2}\widehat i + {y_2}\widehat j + {z_2}\widehat k is given as; p→.q→=(x1×x2)i^+(y1×y2)j^+(z1×z2)k^\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 p→=xi^+yj^+zk^\overrightarrow p = x\widehat i + y\widehat j + z\widehat k is given as; ∣p→∣=x2+y2+z2\left| {\overrightarrow p } \right| = \sqrt {{x^2} + {y^2} + {z^2}}
So, sin⁡θ=∣i^×2i^+0j^×(−3j^)+0k^×6k^∣1(2)2+(3)2+(6)2\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⁡θ=∣2i^∣1(2)2+(3)2+(6)2\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⁡θ=(2)27=27\sin \theta = \dfrac{{\sqrt {{{\left( 2 \right)}^2}} }}{7} = \dfrac{2}{7}or θ=sin⁡−1(27)\theta = {\sin ^{ - 1}}\left( {\dfrac{2}{7}} \right)
Since plane makes an angle of sin⁡−1(α){\sin ^{ - 1}}\left( \alpha \right) with X-axis
Therefore, θ=sin⁡−1(α)\theta = {\sin ^{ - 1}}\left( \alpha \right)
Now substituting the value in the above equation, we get
sin⁡−1(27)=sin⁡−1(α){\sin ^{ - 1}}\left( {\dfrac{2}{7}} \right) = {\sin ^{ - 1}}\left( \alpha \right)
⇒\Rightarrow α=27\alpha = \dfrac{2}{7}
Therefore, the value of α\alpha is equal to 27\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 ∣PQ∣\left| {PQ} \right| .