Question

Find the vector equation of the plane passing through the points (1, 1, −1), (6, 4, −5) and (−4, −2, 3).

Answer 1

 

\[ \text{ Let } A(1, 1, -1),B(6, 4, -5) \text{ and } C(-4, -2, 3).\]
\[\text{ The required plane passes through the point } A(1, 1, -1)  \text{ whose position vector is  }  \vec{a} = \hat{i} + \hat{j}  - \hat{k}  \text{ and is normal to the vector }  \vec{n} \text{ given by } \]
\[ \vec{n} = \vec{AB} \times \vec{AC} \]
\[ \text{ Clearly } , \vec{AB} = \vec{OB} - \vec{OA} = \left( 6 \hat{i} + 4 \hat{j}  - 5 \hat{k}  \right) - \left( \hat{i}  + \hat{j}  - \hat{k}  \right) = 5 \hat{i}  + 3 \hat{j}  - 4 \hat{k} \]
\[ \vec{AC} = \vec{OC} - \vec{OA} = \left( - 4 \hat{i}  - 2 \hat{j}  + 3 \hat{k}  \right) - \left( \hat{i}  + \hat{j}  - \hat{k}  \right) = - 5 \hat{i}  - 3 \hat{j}  + 4 \hat{k}  \]
\[ \vec{n} = \vec{AB} \ × \vec{AC} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k}  \\ 5 & 3 & - 4 \\ - 5 & - 3 & 4\end{vmatrix} = 0 \hat{i}  + 0 \hat{j}  + 0 \hat{k}  = \vec{0} \]
\[\text{ So, the given points are collinear } .\]
\[\text{ Thus, there will be infinite number of planes passing through these points.} \]
\[\text{ Their equations (passing through (1, 1, -1) are given by } \]
\[a \left( x - 1 \right) + b \left( y - 1 \right) + c \left( z + 1 \right) = 0 . . . \left( 1 \right)\]
\[\text{ Since this passes through B }(6, 4, -5),\]
\[a \left( 6 - 1 \right) + b \left( 4 - 1 \right) + c \left( - 5 + 1 \right) = 0\]
\[ \Rightarrow 5a + 3b - 4c = 0 . . . \left( 2 \right)\]
\[ \text{ From (1) and (2), the equations of the infinite planes are } \]
\[a \left( x - 1 \right) + b \left( y - 1 \right) + c \left( z + 1 \right) = 0, \text{ where } 5a + 3b - 4c = 0 . \]