Question

Find the vector equation (in scalar product form) of the plane containing the line of intersection of the planes x − 3y + 2z − 5 = 0 and 2x − y + 3z − 1 = 0 and passing through (1, −2, 3).

Answer 1

\[\text{ The equation of the plane passing through the line of intersection of the given planes is} \]
\[x - 3y + 2z - 5 + \lambda \left( 2x - y + 3z - 1 \right) = 0 . . . \left( 1 \right)\]
\[\text{ This passes through (1, -2, 3). So } ,\]
\[1 + 6 + 6 - 5 + \lambda \left( 2 + 2 + 9 - 1 \right)\]
\[ \Rightarrow 8 + 12\lambda = 0\]
\[ \Rightarrow \lambda = \frac{- 2}{3}\]
\[\text{ Substituting this in (1), we get } \]
\[x - 3y + 2z - 5 - \frac{2}{3} \left( 2x - y + 3z - 1 \right) = 0\]
\[ \Rightarrow - x - 7y - 13 = 0\]
\[ \Rightarrow x + 7y + 13 = 0\]
\[ \Rightarrow \vec{r} . \left( \hat{i}  + 7 \hat{j}  \right) + 13 = 0, \text{ which is the required vector equation of the plane } .\]