Find the angle between the planes $\overline{r}\cdot (2\overline{i}-\overline{j}+2\overline{k})=3$ and $\overline{r}\cdot (3\overline{i}+6\overline{j}+\overline{k})=4$

Given planes are $\overline{r}\cdot (2\overline{i}-\overline{j}+2\overline{k})=3...........\left(1\right)$

$\overline{r}\cdot (3\overline{i}+6\overline{j}+\overline{k})=4\dots ...\left(2\right)$

Eq.(1) is in the form $\overline{r}\cdot \overline{m_1}=p_1$

where $\overline{m_1}=2\overline{i}-\overline{j}+2\overline{k};p_1=3$

Eq.(2) is in the form $\overline{r}\cdot \overline{m_2}=p_2$

where $\overline{m_2}=3\overline{i}+6\overline{j}+\overline{k};p_2=4$

Let $\mathrm{\theta }$ be angle between (1) &(2)

$\therefore \cos \theta =\frac{\overline{m_1}\cdot \overline{m_2}}{\left|\overline{m_1}\right|\left|\overline{m_2}\right|}$

$=\frac{(2\overline{i}-\overline{j}+2\overline{k})\cdot (3\overline{i}+6\overline{j}+\overline{k})}{\sqrt{(2{)}^2+(-1{)}^2+(2{)}^2}\sqrt{(3{)}^2+(6{)}^2+(1{)}^2}}$

$\Rightarrow \cos \theta =\frac{6-6+2}{\sqrt{4+1+4}\sqrt{9+36+1}}=\frac{2}{\sqrt{9}\sqrt{46}}$

$\Rightarrow \cos \theta =\frac{2}{3\sqrt{46}}\Rightarrow \theta ={\cos }^{-1}\left(\frac{2}{3\sqrt{46}}\right)$