Find the general form of the equation of the plane passing through the point and perpendicular to the specified vector or line.
Point: $( 3,0,0 )$ Perpendicular to: $x = 3 - t , y = 2 - 6 t , z = 4 + t$
A) $- x - 6 y - z + 3 = 0$
B) $x - 6 y + z + 3 = 0$
C) $- x + 6 y + z + 3 = 0$
D) $- x - 6 y + z + 3 = 0$
E) $- x - 6 y - z - 3 = 0$

Question

Find the general form of the equation of the plane passing through the point and perpendicular to the specified vector or line. 
Point: $( 3,0,0 )$ Perpendicular to: $x = 3 - t , y = 2 - 6 t , z = 4 + t$ 
A) $- x - 6 y - z + 3 = 0$
B) $x - 6 y + z + 3 = 0$
C) $- x + 6 y + z + 3 = 0$
D) $- x - 6 y + z + 3 = 0$
E) $- x - 6 y - z - 3 = 0$

Quizplus · Accepted Answer

The direction vector of the line is given by the coefficients of $t$ in the parametric equations, which are $(-1, -6, 1)$. This vector is also the normal vector of the plane. The equation of a plane given a point $(x_0, y_0, z_0)$ and a normal vector $(a, b, c)$ is $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$. Substituting the given point $(3, 0, 0)$ and the normal vector $(-1, -6, 1)$, we get $-1(x - 3) - 6(y - 0) + 1(z - 0) = 0$, which simplifies to $-x - 6y + z + 3 = 0$.

Find the General Form of the Equation of the Plane (3,0,0)( 3,0,0 )(3,0,0)

Find the General Form of the Equation of the Plane $( 3,0,0 )$