Find the general form of the equation of the plane passing through the point and perpendicular to the specified line.
$x = - 1 + 5 t$
$( 6 , - 8,9 ) , \quad y = - 3 + t$
$z = - 2 - 6 t$
A) $5 x + y - 6 z = 0$
B) $5 x + y - 6 z - 32 = 0$
C) $5 x + y - 6 z + 32 = 0$
D) $6 x - 8 y + 9 z - 32 = 0$
E) $6 x - 8 y + 9 z + 32 = 0$

Question

Quizplus · Accepted Answer

The direction vector of the line is given by the coefficients of $t$ in the parametric equations, which are $5, 1, -6$. This vector is also normal to the plane. The point $(6, -8, 9)$ lies on the plane. The general form of the equation of a plane is $Ax + By + Cz + D = 0$, where $A, B, C$ are the components of the normal vector, and $D$ can be found by substituting a point on the plane. Substituting $A = 5$, $B = 1$, $C = -6$, and the point $(6, -8, 9)$ into the equation gives $5(6) + 1(-8) - 6(9) + D = 0$, solving for $D$ gives $D = 32$. Therefore, the equation of the plane is $5x + y - 6z + 32 = 0$.

Find the General Form of the Equation of the Plane x=−1+5tx = - 1 + 5 tx=−1+5t

Find the General Form of the Equation of the Plane $x = - 1 + 5 t$