Question
What is the acute angle between the planes $$x + y + 2z = 3$$ and $$ - 2x + y - z = 11\,?$$
A.
$$\frac{\pi }{5}$$
B.
$$\frac{\pi }{4}$$
C.
$$\frac{\pi }{6}$$
D.
$$\frac{\pi }{3}$$
Answer :
$$\frac{\pi }{3}$$
Solution :
The given equation of the planes are $$x + y + 2z = 3$$ and $$ - 2x + y - z = 11$$
We know that, the angle between the planes
$$\eqalign{
& {a_1}x + {b_1}y + {c_1}z + {d_1} = 0{\text{ and}} \cr
& {a_2}x + {b_2}y + {c_2}z + {d_2} = 0{\text{ is given by}} \cr
& \cos \,\theta = \left| {\frac{{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}}}{{\sqrt {a_1^2 + b_1^2 + c_1^2} \,\sqrt {a_2^2 + b_2^2 + c_2^2} }}} \right| \cr
& {\text{Here, }}{a_1} = 1,\,{b_1} = 1,\,{c_1} = 2,\,{a_2} = - 2,\,{b_2} = 1,\,{c_2} = - 1 \cr
& \therefore \,\cos \,\theta = \left| {\frac{{1 \times \left( { - 2} \right) + 1 \times 1 + 2 \times \left( { - 1} \right)}}{{\sqrt {1 + 1 + 4} \,\sqrt {4 + 1 + 1} }}} \right| \cr
& \Rightarrow \cos \,\theta = \left| {\frac{{ - 2 + 1 - 2}}{{\sqrt 6 \,\sqrt 6 }}} \right| \cr
& \Rightarrow \cos \,\theta = \left| {\frac{3}{6}} \right| \cr
& \Rightarrow \cos \,\theta = \frac{1}{2} \cr
& \Rightarrow \cos \,\theta = \cos \frac{\pi }{3} \cr
& \Rightarrow \theta = \frac{\pi }{3}\, \cr} $$