Find the general form of the equation of the plane passing through the point and perpendicular to the specified line.[Be sure to reduce the coefficients in your answer to lowest terms by dividing out any common factor.] $\begin{array} { l }
( - 3,3 , - 7 ) \
x = 4 - t \
y = 1 - 3 t \
z = 3 - 5 t
\end{array}$
A)x + 3y + 5z = 0
B)x + 3y + 5z - 29 = 0
C)x + 3y + 5z + 29 = 0
D)3x - 3y + 7z - 29 = 0
E)3x - 3y + 7z + 29 = 0

Question

Find the general form of the equation of the plane passing through the point and perpendicular to the specified line.[Be sure to reduce the coefficients in your answer to lowest terms by dividing out any common factor.] $\begin{array} { l } 
( - 3,3 , - 7 ) \
x = 4 - t \
y = 1 - 3 t \
z = 3 - 5 t
\end{array}$
A)x + 3y + 5z = 0
B)x + 3y + 5z - 29 = 0
C)x + 3y + 5z + 29 = 0
D)3x - 3y + 7z - 29 = 0
E)3x - 3y + 7z + 29 = 0

Quizplus · Accepted Answer

The direction vector of the line is given by the coefficients of $t$ in the parametric equations, which are $(-1, -3, -5)$. This vector is also the normal vector of the plane. The equation of a plane is given by $Ax + By + Cz + D = 0$, where $A$, $B$, and $C$ are the components of the normal vector, and $D$ can be found by substituting a point through which the plane passes. Substituting the point $(-3, 3, -7)$ into the equation $x + 3y + 5z + D = 0$ gives $(-3) + 3(3) + 5(-7) + D = 0$, which simplifies to $-3 + 9 - 35 + D = 0$, leading to $D = 29$. Therefore, the equation of the plane is $x + 3y + 5z + 29 = 0$.

Find the General Form of the Equation of the Plane (−3,3,−7)x=4−ty=1−3tz=3−5t\begin{array} { l } ( - 3,3 , - 7 ) \\x = 4 - t \\y = 1 - 3 t \\z = 3 - 5 t\end{array}(−3,3,−7)x=4−ty=1−3tz=3−5t​

Find the General Form of the Equation of the Plane $\begin{array} { l } ( - 3,3 , - 7 ) \\x = 4 - t \\y = 1 - 3 t \\z = 3 - 5 t\end{array}$