Question

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

Answer 1

Given:

$\overrightarrow{r}.(2\hat{i}+2\hat{j}-3\hat{k})=5$ and $\overrightarrow{r}.(3\hat{i}-3\hat{j}+5\hat{k})=3$

$\Rightarrow \overrightarrow{n_1}=2\hat{i}+2\hat{j}-3\hat{k};\overrightarrow{n_2}=3\hat{i}-3\hat{j}+5\hat{k}$

If $\theta$ is the angle between the planes, then

$\cos \theta \left|\frac{\overrightarrow{n_1}.\overrightarrow{n_2}}{|\overrightarrow{n_1}||\overrightarrow{n_2}|}\right|=\left|\frac{(2\hat{i}+2\hat{j}-3\hat{k})\cdot (3\hat{i}-3\hat{j}+5\hat{k})}{\sqrt{4+4+9}\sqrt{9+9+25}}\right|$

$\cos \theta =\left|\frac{6-6-15}{\sqrt{17}\sqrt{43}}\right|=\frac{15}{\sqrt{731}}$

or $\theta ={\cos }^{-1}\left(\frac{15}{\sqrt{731}}\right)$.