Question

Find the angles between the plane $\overrightarrow{r}.(\hat{i}-2\hat{j}-2\hat{k})=1$ and $\overrightarrow{r}.(3\hat{i}-6\hat{j}+2\hat{k})=0$.

Answer 1

The angle between two planes is the angle between the normal to the two planes.$\theta ={\cos }^{-1}\left(\frac{\overrightarrow{n_1}.\overrightarrow{n_2}}{\left|\overrightarrow{n_1}\right|\left|\overrightarrow{n_2}\right|}\right)$

If $\overrightarrow{n_1}.\overrightarrow{n_2}=0$ the planes are at right angles.

$\begin{array}{c}\overrightarrow{r}\cdot \left(\hat{i}-2\hat{j}-2\hat{k}\right)=1\overrightarrow{r}\cdot \left(3\hat{i}-6\hat{j}+2\hat{k}\right)=0\\ \overrightarrow{n_1}=\hat{i}-2\hat{j}-2\hat{k},\overrightarrow{n_2}=3\hat{i}-6\hat{j}+2\hat{k}arethenormalvectors\\ \theta ={\cos }^{-1}\left(\frac{\overrightarrow{n_1}.\overrightarrow{n_2}}{\left|\overrightarrow{n_1}\right|\left|\overrightarrow{n_2}\right|}\right)={\cos }^{-1}\left(\frac{\left(1\right)\left(3\right)+(-2)(-6)+(-2)\left(2\right)}{\sqrt{1+4+4}\sqrt{9+36+4}}\right)\\ ={\cos }^{-1}\frac{\left(3+12-4\right)}{\sqrt{441}}=\frac{11}{21}\end{array}$