Find the general form of the equation of the plane with the given characteristics.
The plane passes through the points $( - 4,4,3 )$ and $( 2,7 , - 7 )$ and is perpendicular to the plane $2 x + 3 y + 3 z = 1$.
A) $39 x - 38 y + 12 z + 272 = 0$
B) $2 x + 3 y + 3 z + 13 = 0$
C) $39 x + 38 y + 12 z + 32 = 0$
D) $2 x + 3 y + 3 z = 0$
E) $38 x - 39 y + 14 z + 266 = 0$

Question

Quizplus · Accepted Answer

The normal vector of the plane we're looking for must be parallel to the normal vector of the given plane $2x + 3y + 3z = 1$, which is $\langle 2, 3, 3 \rangle$, and also perpendicular to the vector formed by the two given points. The vector between the points $(-4,4,3)$ and $(2,7,-7)$ is $\langle 6, 3, -10 \rangle$. The cross product of $\langle 2, 3, 3 \rangle$ and $\langle 6, 3, -10 \rangle$ gives the normal vector of the plane we're looking for, which is $\langle 39, -38, 12 \rangle$. Using one of the points to solve for the equation of the plane, we get $39x - 38y + 12z + D = 0$, where $D$ is determined by plugging in one of the points. The correct equation that fits this description is $39x - 38y + 12z + 272 = 0$.

Find the General Form of the Equation of the Plane (−4,4,3)( - 4,4,3 )(−4,4,3)

Find the General Form of the Equation of the Plane $( - 4,4,3 )$