Question

Find the vector equation of the plane passing through the points \[3 \hat{i}  + 4 \hat{j}  + 2 \hat{k} , 2 \hat{i} - 2 \hat{j} - \hat{k}  \text{ and }  7 \hat{i}  + 6 \hat{k}  .\]

 

Answer 1

\[ \text{ Let } A(3, 4, 2),B(2, -2, -1) \text{ and } C(7, 0, 6) \text{ be the points represented by the given position vectors } .\]
\[\text{ The required plane passes through the point A (3, 4, 2) whose position vector is } \vec{a} =3 \text{i} +4 \hat{j} +2 \hat{k}  \text{ and is normal to the vector } \vec{n} \text{ given by } \]
\[ \vec{n} = \vec{AB} \times \vec{AC} . \]
\[ \text{ Clearly } , \vec{AB} = \vec{OB} - \vec{OA} = \left( 2 \hat{i}  - 2 \hat{j} - \hat{k} \right) - \left( 3 \hat{i} +4 \hat{j} +2 \hat{k} \right) = - \hat{i}  - 6 \hat{j}  - 3 \hat{k}  \]
\[ \vec{AC} = \vec{OC} - \vec{OA} = \left( 7 \hat{i} + 0 \hat{j}  + 6 \hat{k}  \right) - \left( 3 \hat{i}  +4 \hat{j}  +2 \hat{k}  \right) = 4 \hat{i}  - 4 \hat{j}  + 4 \hat{k}  \]
\[ \vec{n} = \vec{AB} \ ×  \vec{AC} = \begin{vmatrix}\hat{i} & \hat{j}  & \hat{k}  \\ - 1 & - 6 & - 3 \\ 4 & - 4 & 4\end{vmatrix} = - 36 \hat{i}  - 8 \hat{j}  + 28 \hat{k} \]
\[ \text{ The vector equation of the required plane is } \]
\[ \vec{r} . \vec{n} = \vec{a} . \vec{n} \]
\[ \Rightarrow \vec{r} . \left( - 36 \hat{i} - 8 \hat{j}  + 28 \hat{k}  \right) = \left( 3 \hat{i}  +4 \hat{j}  +2 \hat{k}  \right) . \left( - 36 \hat{i}   - 8 \hat{j}  + 28 \hat{k}  \right)\]
\[ \Rightarrow \vec{r} . \left[ - 4 \left( 9 \hat{i}  + 2 \hat{j} + 7 \hat{k}  \right) \right] = - 108 - 32 + 56\]
\[ \Rightarrow \vec{r} . \left[ - 4 \left( 9 \hat{i}  + 2 \hat{j}  + 7 \hat{k}  \right) \right] = - 84\]
\[ \Rightarrow \vec{r} . \left( 9 \hat{i}  + 2 \hat{j}  + 7 \hat{k}  \right) = 21\]