Question

Find the measure of angle between the planes \[3x - 4y + 5z = 0\] and \[2x - y - 2z = 5\].
A. \[\dfrac{\pi }{3}\]
B. \[\dfrac{\pi }{2}\]
C. \[\dfrac{\pi }{6}\]
D. None of these

Answer 1

Hint: First, compare the given equations of planes with the general equation of plain and calculate the values of the coefficients. Then, substitute the values in the formula of the angle between the two planes and simplify it. In the end, calculate the value of \[\theta \] and get the required answer.



Formula Used:The angle between the two planes \[{a_1}x + {b_1}y + {c_1}z + {d_1} = 0\] and \[{a_2}x + {b_2}y + {c_2}z + {d_2} = 0\] is:
\[\cos \theta = \left| {\dfrac{{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}}}{{\sqrt {{a_1}^2 + {b_1}^2 + {c_1}^2} \sqrt {{a_2}^2 + {b_2}^2 + {c_2}^2} }}} \right|\]



Complete step by step solution:The given equations of the planes are:
\[3x - 4y + 5z = 0\] \[.....\left( 1 \right)\]
\[2x - y - 2z = 5\]
\[ \Rightarrow 2x - y - 2z - 5 = 0\] \[.....\left( 2 \right)\]

Now compare the equations \[\left( 1 \right)\] and \[\left( 2 \right)\] with \[{a_1}x + {b_1}y + {c_1}z + {d_1} = 0\] and \[{a_2}x + {b_2}y + {c_2}z + {d_2} = 0\].
We get,
\[{a_1} = 3,{b_1} = - 4,{c_1} = 5,{d_1} = 0\] and \[{a_2} = 2,{b_2} = - 1,{c_2} = - 2,{d_2} = - 5\]
Now apply the formula of the angle between the two planes.
\[\cos \theta = \left| {\dfrac{{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}}}{{\sqrt {{a_1}^2 + {b_1}^2 + {c_1}^2} \sqrt {{a_2}^2 + {b_2}^2 + {c_2}^2} }}} \right|\]
Substitute the values in the formula.
\[\cos \theta = \left| {\dfrac{{\left( 3 \right)\left( 2 \right) + \left( { - 4} \right)\left( { - 1} \right) + \left( 5 \right)\left( { - 2} \right)}}{{\sqrt {{3^2} + {{\left( { - 4} \right)}^2} + {5^2}} \sqrt {{2^2} + {{\left( { - 1} \right)}^2} + {{\left( { - 2} \right)}^2}} }}} \right|\]
\[ \Rightarrow \cos \theta = \left| {\dfrac{{6 + 4 - 10}}{{\sqrt {9 + 16 + 25} \sqrt {4 + 1 + 4} }}} \right|\]
\[ \Rightarrow \cos \theta = \dfrac{0}{{\sqrt {50} \sqrt 9 }}\]
\[ \Rightarrow \cos \theta = 0\]
\[ \Rightarrow \theta = {\cos ^{ - 1}}\left( 0 \right)\]
\[ \Rightarrow \theta = \dfrac{\pi }{2}\]
Thus, the angle between the planes \[3x - 4y + 5z = 0\] and \[2x - y - 2z = 5\] is \[\dfrac{\pi }{2}\].



Option ‘B’ is correct



Note: Sometimes students use the formula of the angle between the planes as \[\cos \theta = \dfrac{{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}}}{{\sqrt {{a_1}^2 + {b_1}^2 + {c_1}^2} \sqrt {{a_2}^2 + {b_2}^2 + {c_2}^2} }}\]. If the value of \[{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2} < 0\], then they will get different value. So, always remember to apply the modulus function.