Question

Find the direction cosines of the unit vector perpendicular to the plane  \[\vec{r} \cdot \left( 6 \hat{i}  - 3 \hat{j} - 2 \hat{k} \right) + 1 = 0\] 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.

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