Find the angle between the planes whose vector equations are $\vec r \cdot (2\hat i + 2\hat j - 3\hat k) = 5$ and $\vec r \cdot (3\hat i - 3\hat j + 5\hat k) = 3.$

If $\theta$ be the angle between the planes

$\vec r \cdot {\vec n_1} = {d_1}$ and $\vec r \cdot {\vec n_2} = {d_2},$ then $\cos \theta = \left| {\cfrac{{{{\vec n}_1} \cdot {{\vec n}_2}}}{{|{{\vec n}_1}||{{\vec n}_2}|}}} \right|$

Here ${\vec n_1} = 2\hat i + 2\hat j - 3\hat k$ and ${\vec n_2} = 3\hat i - 3\hat j + 5\hat k$

$\cos \theta = \left| {\cfrac{{(2\hat i + 2\hat j - 3\hat k) \cdot (3\hat i - 3\hat j + 5\hat k)}}{{\sqrt {4 + 4 + 9} \sqrt {9 + 9 + 25} }}} \right|$ $= \left| {\cfrac{{6 - 6 - 15}}{{\sqrt {17} \sqrt {13} }}} \right| = \left| {\cfrac{{ - 15}}{{\sqrt {17} \sqrt {13} }}} \right|$

Here, $\theta = {\cos ^{ - 1}}\left| {\cfrac{{ - 15}}{{\sqrt {17} \sqrt {43} }}} \right| = {\cos ^{ - 1}}\left( {\cfrac{{15}}{{\sqrt {731} }}} \right)$