Question

Find the direction cosines of the unit vector perpendicular to the plane $\overrightarrow{r}.(6\hat{i}-3\hat{j}-2k)+1=0$ passing through the origin.

Answer 1

The equation of the plane is $\overrightarrow{r}.(6\hat{i}-3\hat{j}-2\hat{k})+1=0$. To find the direction cosines of the normal through the origin, we must write the equation in normal form as follows.
$\overrightarrow{r}.(6\hat{i}-3\hat{j}-2\hat{k})+1=0$
$\Rightarrow \overrightarrow{r}.(-6\hat{i}+3\hat{j}+2\hat{k})=1$
$\Rightarrow \overrightarrow{r}.\frac{(-6\hat{i}+3\hat{j}+2\hat{k})}{\sqrt{36+9+4}}=\frac{1}{\sqrt{36+9+4}}\Rightarrow \overrightarrow{r}(-\frac{6}{7}\hat{i}+\frac{3}{7}\hat{j}+\frac{2}{7}\hat{k})=\frac{1}{7}$
Hence,m direction cosines of the normal vector through the origin are $-\frac{6}{7},\frac{3}{7},\frac{2}{7}$.