Question

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

Answer 1

For the unit vector perpendicular to the given plane, we need to convert the given equation of plane into normal form.

$$\begin{gathered} \text{The given equation of the plane is} \\ \overrightarrow{r}.\left(6\hat{i}-3\hat{j}-2\hat{k}\right)+1=0 \\ \Rightarrow \overrightarrow{r}.\left(6\hat{i}-3\hat{j}-2\hat{k}\right)=-1 \\ \Rightarrow \overrightarrow{r}.\left(-6\hat{i}+3\hat{j}+2\hat{k}\right)=1...\left(1\right) \\ \text{Now,}\sqrt{{\left(-6\right)}^2+3^2+2^2}\text{=}\sqrt{36+9+4}\text{= 7} \\ \text{Dividing (1) by 7, we get} \\ \overrightarrow{r}.\left(\frac{-6}{7}\hat{i}+\frac{3}{7}\hat{j}+\frac{2}{7}\hat{k}\right)=\frac{1}{7},\text{which is in the normal form}\overrightarrow{r}\text{.}\overrightarrow{n}\text{=}d, \\ \text{where the unit vector normal to the given plane,}\overrightarrow{n}=\frac{-6}{7}\hat{i}+\frac{3}{7}\hat{j}+\frac{2}{7}\hat{k} \\ \text{So, its direction cosines are}\frac{-6}{7}\text{,}\frac{3}{7}\text{,}\frac{2}{7} \end{gathered}$$