Question

The angle between two planes $x + 2y + 2z = 3$ and $- 5x + 3y + 4z = 9$ is
A. ${\cos ^{ - 1}}\dfrac{{9\sqrt 2 }}{{10}}$
B. ${\cos ^{ - 1}}\dfrac{{3\sqrt 2 }}{5}$
C. ${\cos ^{ - 1}}\dfrac{{3\sqrt 2 }}{{10}}$
D. ${\cos ^{ - 1}}\dfrac{{19\sqrt 2 }}{{30}}$

Answer 1

Hint : We are given two equations of the plane in the cartesian coordinate system that is in x ,y and z form. First we will write them in standard form. That is in the form of $Ax + By + Cz + D = 0$ . In this question we have two planes given and we have to find the angle between them. So will simply use the formula to find the angle.
$\cos \theta = \left| {\dfrac{{{A_1}{A_2} + {B_1}{B_2} + {C_1}{C_2}}}{{\sqrt {{A_1}^2 + {B_1}^2 + {C_1}^2} \sqrt {{A_2}^2 + {B_2}^2 + {C_2}^2} }}} \right|$

Complete step by step solution:
Given that two planes are $x + 2y + 2z = 3$ and $- 5x + 3y + 4z = 9$
Let first write them in standard form.
So the equations are $x + 2y + 2z - 3 = 0\& \- 5x + 3y + 4z - 9 = 0$
Let the equations as ${A_1}x + {B_1}y + {C_1}z + {D_1} = 0\& {A_2}x + {B_2}y + {C_2}z + {D_2} = 0$ respectively.
Now just substituting the values we get,
$\cos \theta = \left| {\dfrac{{1\left( { - 5} \right) + 2 \times 3 + 2 \times 4}}{{\sqrt {{1^2} + {2^2} + {2^2}} \sqrt {{{\left( { - 5} \right)}^2} + {3^2} + {4^2}} }}} \right|$
Now on solving the roots we get,
$\cos \theta = \left| {\dfrac{{ - 5 + 6 + 8}}{{\sqrt {1 + 4 + 4} \sqrt {25 + 9 + 16} }}} \right|$
Taking the sum of numbers inside the roots,
$\cos \theta = \left| {\dfrac{9}{{\sqrt 9 \sqrt {50} }}} \right|$
Now we will take the perfect roots out,
$\cos \theta = \left| {\dfrac{9}{{3 \times 5\sqrt 2 }}} \right|$
On dividing by 3 we get,
$\cos \theta = \left| {\dfrac{3}{{5\sqrt 2 }}} \right|$
This is the answer. But this is not available in the option. So we need to modify this. Now multiply numerator and denominator by root 2.
$\cos \theta = \left| {\dfrac{{3\sqrt 2 }}{{5\sqrt 2 \times \sqrt 2 }}} \right|$
The denominator will be,
$\cos \theta = \left| {\dfrac{{3\sqrt 2 }}{{5 \times 2}}} \right|$
$\cos \theta = \left| {\dfrac{{3\sqrt 2 }}{{10}}} \right|$
On taking the modulus,
$\cos \theta = \dfrac{{3\sqrt 2 }}{{10}}$
To find the angle we can write,
$\theta = {\cos ^{ - 1}}\dfrac{{3\sqrt 2 }}{{10}}$
This is the available answer. Thus option C is the correct answer.
So, the correct answer is “Option C”.

Note : Here note the options are confusing. Sometimes they can give two correct options but there can be some modification in writing the way. As we observed above.
Also note that the modulus is always a positive value though the value inside is negative