Question

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

Answer 1

The equations of the given planes are $\overrightarrow{r}.(2\hat{i}+2\hat{j}-3\hat{k})=5$ and $\overrightarrow{r}.(3\hat{i}-3\hat{j}+5\hat{k})$
It is known that if $\overrightarrow{n_1}$ and $\overrightarrow{n_2}$ are normal to the planes $\overrightarrow{r}.\overrightarrow{n_1}=d_1$ and $\overrightarrow{r}.\overrightarrow{n_2}=d_2$ then the angle between them, $\theta$ is given by
$\cos \theta =\left|\frac{\overrightarrow{n_1}.\overrightarrow{n_2}}{\left|\overrightarrow{n_1}\right|\left|\overrightarrow{n_2}\right|}\right|....\left(1\right)$
Here $\overrightarrow{n_1}=2\hat{i}+2\hat{j}-3\hat{k}$ and $\overrightarrow{n_2}=3\hat{i}-3\hat{j}+5\hat{k}$
$\therefore$ $\overrightarrow{n_1}.\overrightarrow{n_2}=(2\hat{i}+2\hat{j}-3\hat{k})(3\hat{i}-3\hat{j}+5\hat{k})=2.3+2.(-3)+(-3).5=-15$
$\left|\overrightarrow{n_1}\right|=\sqrt{{\left(2\right)}^2+{\left(2\right)}^2+{(-3)}^2}=\sqrt{17}$
$\left|\overrightarrow{n_2}\right|=\sqrt{{\left(3\right)}^2+{(-3)}^2+{\left(5\right)}^2}=\sqrt{43}$
Substituting the value of $\overrightarrow{n_1}.\overrightarrow{n_2}$ and $\left|\overrightarrow{n_1}\right|\left|\overrightarrow{n_2}\right|$ in equation (1), we obtian
$\cos \theta =\left|\frac{-15}{\sqrt{17}.\sqrt{43}}\right|$
$\Rightarrow$ $\cos \theta =\frac{15}{\sqrt{731}}$
$\Rightarrow \theta =$ ${\cos }^{-1}\frac{15}{\sqrt{731}}$