Question

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

x − y + z − 2 = 0 and 3x + 2y − z + 4 = 0 

Answer 1

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