Question
Find the angle between planes $\bar{r} \cdot(\hat{i}+\hat{j}-2 \hat{k})=8$ and $\bar{r} \cdot(-2 \hat{i}+\hat{j}+\hat{k})=3$

Answer

Normal to the given planes are $\bar{n}_1=\hat{i}+\hat{j}-2 \hat{k}$ and $\bar{n}_2=-2 \hat{i}+\hat{j}+\hat{k}$
The acute angle $\theta$ between normal is given by
$
\begin{aligned}
& \cos \theta=\left|\frac{\bar{n}_1 \cdot \bar{n}_2}{\left|\bar{n}_1\right| \cdot\left|\bar{n}_2\right|}\right| \\
& \therefore \cos \theta=\left|\frac{\mid(\hat{i}+\hat{j}-2 \hat{k}) \cdot(-2 \hat{i}+\hat{j}+\hat{k})}{\sqrt{6} \cdot \sqrt{6}}\right|=\left|\frac{-3}{6}\right|=\frac{1}{2} \\
& \therefore \cos \theta=\frac{1}{2} \quad \therefore \theta=60^{\circ}=\frac{\pi}{3}
\end{aligned}
$
The acute angle between normals $\bar{n}_1$ and $\bar{n}_2$ is $60^{\circ}$.
$\therefore \quad$ The angle between given planes is $60^{\circ}$.

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