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

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.