Find the general form of the equation of the plane passing through the point and perpendicular to the specified line.
$x = 2 + 3 t$
$( 5 , - 6 , - 4 ) , \quad y = - 3 + 4 t$
$z = - 6 + 6 t$
A) $3 x + 4 y + 6 z = 0$
B) $3 x + 4 y + 6 z - 33 = 0$
C) $3 x + 4 y + 6 z + 33 = 0$
D) $5 x - 6 y - 4 z - 33 = 0$
E) $5 x - 6 y - 4 z + 33 = 0$

Question

Quizplus · Accepted Answer

The normal vector of the plane is parallel to the direction vector of the line, which is $\langle 3, 4, 6 \rangle$. So the equation of the plane is of the form $3x + 4y + 6z = d$. To find the value of $d$, we substitute the coordinates of the given point into the equation:$$3(5) + 4(-6) + 6(-4) = d$$$$15 - 24 - 24 = d$$$$-33 = d$$So the equation of the plane is $3x + 4y + 6z = -33$.

Find the General Form of the Equation of the Plane x=2+3tx = 2 + 3 tx=2+3t

Find the General Form of the Equation of the Plane $x = 2 + 3 t$