Question

find the angles between the planes x+2y+2z-5=0and 3x+3y+2z-8=0

Answer 1

Angle between two planes is same as angle between the vectors normal to them.

vector normal to the first plane is given by :
         $\vec{A} = i + 2 j + 2 k\\\\magnitude = \sqrt{1^2+2^2+2^2}=3$

vector normal to the second plane is given by
       $\vec{B}= 3 i + 3 j + 2 k \\\\magnitude=\sqrt{3^2+3^2+2^2}=\sqrt{22}$
          
Now, angle  between the two vectors  be  Ф.  The dot product of vectors is:

$\vec{A}.\vec{B}=A * B * Cos\phi\\\\(i +2 j + 2k) . (3i+3j+2k)=A * B * Cos\phi\\\\3*\sqrt{22}*Cos\phi=3+6+4=13\\\\Cos\phi=\frac{13}{3\ \sqrt{22}}\\\\\phi=22.50^0$