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Question
Find the vector equation of the plane through the points (2, 1, −1) and (−1, 3, 4) and perpendicular to the plane x − 2y + 4z = 10
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Solution
\[\text{ The equation of any plane passing through (2, 1, -1) is } \]
\[a \left( x - 2 \right) + b \left( y - 1 \right) + c \left( z + 1 \right) = 0 . . . \left( 1 \right)\]
\[ \text{ It is given that (1) is passing through (-1, 3, 4). So } ,\]
\[a \left( - 1 - 2 \right) + b \left( 3 - 1 \right) + c \left( 4 + 1 \right) = 0\]
\[ \Rightarrow - 3a + 2b + 5c . . . \left( 2 \right)\]
\[\text{ It is given that (1) is perpendicular to the plane x - 2y + 4z = 10 . So } ,\]
\[a - 2b + 4c = 0 . . . \left( 3 \right)\]
\[ \text{ Solving (1), (2) and (3), we get } \]
\[\begin{vmatrix}x - 2 & y - 1 & z + 1 \\ - 3 & 2 & 5 \\ 1 & - 2 & 4\end{vmatrix} = 0\]
\[ \Rightarrow 18 \left( x - 2 \right) + 17 \left( y - 1 \right) + 4 \left( z + 1 \right) = 0\]
\[ \Rightarrow 18x + 17y + 4z - 49 = 0\]
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