Question

Find the angle between planes $\overrightarrow{r}.(\hat{i}-\hat{j}+\hat{k})=5$ and $\overrightarrow{r}.(2\hat{i}+\hat{j}-\hat{k})=7$

Answer 1

$P_1:\overrightarrow{r}.(\hat{i}-\hat{j}+\hat{k})=5$

$P_2:\overrightarrow{r}.(2\hat{i}+\hat{j}-\hat{k})=7$

Normal of plane 1: $n_1:\hat{i}-\hat{j}+\hat{k}$

Normal of plane 2: $n_2:2\hat{i}+\hat{j}-\hat{k}$

Angle between two planes, is same as angle between its normals.

Therefore,

$n_1\cdot n_2=(\hat{i}-\hat{j}+\hat{k})\cdot (2\hat{i}+\hat{j}-\hat{k})$

$|n_1||n_2|\cos \theta =2-1-1$

$\Rightarrow \sqrt{3}\sqrt{6}\cos \theta =0$

$\Rightarrow \cos \theta =0$

$\Rightarrow \theta =90˚$

Therefore, angle between given planes is $90˚$.