Question

Show that the normals to the following pairs of planes are perpendicular to each other.

(i) x − y + z − 2 = 0 and 3x + 2y − z + 4 = 0

(ii) $\overrightarrow{r}·\left(2\hat{i}-\hat{j}+3\hat{k}\right)=5\mathrm{and}\overrightarrow{r}·\left(2\hat{i}-2\hat{j}-2\hat{k}\right)=5$

Answer 1

$$\begin{gathered} \left(i\right)\text{Let}\overrightarrow{n_1}\text{and}\overrightarrow{n_2}\text{be the vectors which are normals to the planes}x-y+z=2\text{and}3x+2y-z=-4\text{respectively.} \\ \text{The given equations of the planes are} \\ x-y+z=2;3x+2y-z=-4 \\ \Rightarrow \left(x\hat{i}+y\hat{j}+z\hat{k}\right).\left(\hat{i}-\hat{j}+\hat{k}\right)=8;\left(x\hat{i}+y\hat{j}+z\hat{k}\right).\left(3\hat{i}+2\hat{j}-\hat{k}\right)=-4 \\ \Rightarrow \overrightarrow{n_1}=\hat{i}-\hat{j}+\hat{k};\overrightarrow{n_2}=3\hat{i}+2\hat{j}-\hat{k} \\ \text{Now,}\overrightarrow{n_1}\text{.}\overrightarrow{n_2}=\left(\hat{i}-\hat{j}+\hat{k}\right).\left(3\hat{i}+2\hat{j}-\hat{k}\right)=3-2-1=0 \\ \text{So, the normals to the given planes are perpendicular to each other.} \end{gathered}$$

$$\begin{gathered} \left(ii\right)\text{Let}\overrightarrow{n_1}\text{and}\overrightarrow{n_2}\text{be the vectors which are normals to the planes}\overrightarrow{r}\text{.}\left(2\hat{i}-\hat{j}+3\hat{k}\right)\text{= 5}\text{and}\overrightarrow{r}\text{.}\left(2\hat{i}-2\hat{j}-2\hat{k}\right)\text{= 5}\text{respectively.} \\ \text{The given equations of the planes are} \\ \overrightarrow{r}\text{.}\left(2\hat{i}-\hat{j}+3\hat{k}\right)\text{= 5};\overrightarrow{r}\text{.}\left(2\hat{i}-2\hat{j}-2\hat{k}\right)\text{= 5} \\ \Rightarrow \overrightarrow{n_1}=\left(2\hat{i}-\hat{j}+3\hat{k}\right);\overrightarrow{n_2}=\left(2\hat{i}-2\hat{j}-2\hat{k}\right) \\ \text{Now,}\overrightarrow{n_1}\text{.}\overrightarrow{n_2}=\left(2\hat{i}-\hat{j}+3\hat{k}\right).\left(2\hat{i}-2\hat{j}-2\hat{k}\right)=4+2-6=0 \\ \text{So, the normals to the given planes are perpendicular to each other.} \end{gathered}$$