Question

Find the equation of the plane through (2, 3, −4) and (1, −1, 3) and parallel to x-axis.

 

Answer 1

\[\text{ The equation of the plane through (2, 3, -4) is } \]
\[a \left( x - 2 \right) + b \left( y - 3 \right) + c \left( z + 4 \right) = 0 . . . \left( 1 \right)\]
\[\text{ This plane passes through (1, -1, 3). So } ,\]
\[a \left( 1 - 2 \right) + b \left( - 1 - 3 \right) + c \left( 3 + 4 \right) = 0\]
\[ \Rightarrow - a - 4b + 7c = 0 . . . \left( 2 \right)\]
\[\text{ Again plane (1) is parallel to x-axis. It means that plane (1) is perpendicular to the yz-plane whose equation is x = 0 or 1 . x + 0 . y + 0 . z = 0 } \]
\[ \Rightarrow a \left( 1 \right) + b \left( 0 \right) + c \left( 0 \right) = 0 . . . \left( 3 \right) (\text{ Because a}_1 a_2 + b_1 b_2 + c_1 c_2 = 0)\]
\[\text{ Solving (1), (2) and (3), we get} \]
\[\begin{vmatrix}x - 3 & y - 3 & z + 4 \\ - 1 & - 4 & 7 \\ 1 & 0 & 0\end{vmatrix} = 0\]
\[ \Rightarrow 0 \left( x - 3 \right) + 7 \left( y - 3 \right) + 4 \left( z + 4 \right) = 0\]
\[ \Rightarrow 7y + 4z - 5 = 0\]