Question
What is the angle between the planes $$2x - y + z = 6$$ and $$x + y + 2z = 3\,?$$
A.
$$\frac{\pi }{2}$$
B.
$$\frac{\pi }{3}$$
C.
$$\frac{\pi }{4}$$
D.
$$\frac{\pi }{6}$$
Answer :
$$\frac{\pi }{3}$$
Solution :
We know, if $${a_1}x + {b_1}y + {c_1}z = {d_1}$$ and $${a_2}x + {b_2}y + {c_2}z = {d_2}$$
are two planes then angle between them is
$$\cos \,\theta = \frac{{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}}}{{\sqrt {a_1^2 + b_1^2 + c_1^2} \,\sqrt {a_2^2 + b_2^2 + c_2^2} }}$$
Let $$q$$ be the angle between given planes. Here,
$$\eqalign{
& {a_1} = 2,\,{b_1} = - 1,\,{c_1} = 1\,;\,{a_2} = 1,\,{b_2} = 1,\,{c_2} = 2 \cr
& \therefore \,\cos \,q = \frac{{2 \times 1 + 1 \times \left( { - 1} \right) + 1 \times 2}}{{\sqrt {4 + 1 + 1} \,\sqrt {1 + 1 + 4} }} \cr
& \Rightarrow \cos \,q = \frac{3}{6} = \frac{1}{2} \cr
& \Rightarrow \cos \,q = \cos \,\frac{\pi }{3} \cr
& \Rightarrow \theta = \frac{\pi }{3} \cr} $$