Question

Find the vector equation of a plane which is at a distance of 5 units from the origin and its normal vector is \[2 \hat{i} - 3 \hat{j} + 6 \hat{k} \] .

Answer 1

Given: 

\[\text{Normal vector } , \hat{n} = 2 \hat{i} - 3 \hat{j} + 6 \hat{k}  \]
\[\text{ Perpendicular distance, d = 5 units } \]

The vector equation of a plane that is at a distance of 5 units from the origin and has its normal vector \[\hat{n} = 2 \hat{i}  - 3 \hat{j}  + 6 \hat{k} \] is as follows:

\[\vec{r .} \hat{n}  = d\]         

  \[\vec{r .} (2 \hat{i} - 3 \hat{j} + 6 \hat{k}  ) = 5\]