Question

The angle between a normal to the plan \[2x - y + 2z - 1 = 0\]and \[Z\]-axis is
A) \[{\cos ^{ - 1}}\left( {\dfrac{1}{3}} \right)\]
B) \[{\sin ^{ - 1}}\left( {\dfrac{2}{3}} \right)\]
C) \[{\cos ^{ - 1}}\left( {\dfrac{2}{3}} \right)\]
D) \[{\sin ^{ - 1}}\left( {\dfrac{1}{3}} \right)\]
E) \[{\sin ^{ - 1}}\left( {\dfrac{3}{5}} \right)\]

Answer 1

Hint: We have given two plane equations so the coefficient of plan equation gives directional ratios of plane respectively. As the given plane meets the \[Z\]-axis put \[x = 0,y = 0\]in the plane equation\[2x - y + 2z - 1 = 0\]so we get value of z after that apply the formula for finding angle between a normal to the plane which is equal to\[\cos \theta = \dfrac{{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}}}{{\sqrt {{a_1}^2 + {b_1}^2 + {c_1}^2} \sqrt {{a_2}^2 + {b_2}^2 + {c_2}^2} }}\]

Complete step by step solution:
Given equation of plane is \[2x - y + 2z - 1 = 0 \ldots \left[ 1 \right]\]
\[\therefore \] Directional ratios of the given the given plane is \[\left( {2, - 1,2} \right)\] and directional ration of the Z-plane is \[\left( {0,0,1} \right)\]
Given a plane, meet the \[Z\]-axis.
\[\therefore \] Put \[x = 0,y = 0\] in equation \[\left[ 1 \right]\]
So we get,
\[2\left( 0 \right) - \left( 0 \right) + 2z - 1 = 0\]
\[ \Rightarrow 2z - 1 = 0\]
\[ \Rightarrow z = \dfrac{1}{2}\]
So, the point on \[Z\]-axis is\[\left( {0,0,\dfrac{1}{2}} \right)\]
The angle between two planes having direction ratios \[{a_1},{b_1},{c_1}\]and \[{a_2},{b_2},{c_2}\] is \[\theta \], then
\[\cos \theta = \dfrac{{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}}}{{\sqrt {{a_1}^2 + {b_1}^2 + {c_1}^2} \sqrt {{a_2}^2 + {b_2}^2 + {c_2}^2} }}\]
Here \[{a_1} = 2\], \[{b_1} = - 1\], \[{c_1} = 2\] and \[{a_2} = 0\], \[{b_2} = 0\], \[{c_2} = 1\]
On substituting the values in given equation we get,
\[ \Rightarrow \] \[\cos \theta = \dfrac{{2 \times 0 - 1 \times 0 + 2 \times 1}}{{\sqrt {{2^2} + {{( - 1)}^2} + {2^2}} \sqrt {{{\left( 0 \right)}^2} + {{(0)}^2} + {{\left( 1 \right)}^2}} }}\]
On simplification we get,
\[ \Rightarrow \cos \theta = \dfrac{2}{{\sqrt {(4 + 1 + 4)} \times \sqrt 1 }}\]
\[ \Rightarrow \cos \theta = \dfrac{2}{{\sqrt 9 \times \sqrt 1 }}\]
As we know \[\sqrt 9 = 3\]and \[\sqrt 1 = 1\],
\[ \Rightarrow \cos \theta = \dfrac{2}{{3 \times 1}}\]
\[ \Rightarrow \cos \theta = \dfrac{2}{3}\]
\[ \Rightarrow \theta = {\cos ^{ - 1}}\left( {\dfrac{2}{3}} \right)\]
Thus, the angle between a normal to the plan \[2x - y + 2z - 1 = 0\] and \[Z\]-axis is\[{\cos ^{ - 1}}\left( {\dfrac{2}{3}} \right)\]
Hence, option c \[{\cos ^{ - 1}}\left( {\dfrac{2}{3}} \right)\] is the correct answer.

Note: The angle between planes is equal to an angle between their normal vectors that is the angle between planes is equal to an angle between lines \[{l_1}\]and \[{l_2}\], which lie on planes and which is perpendicular to lines of planes crossing.