Find an equation of a plane passing through the given point and perpendicular to the given vector. Point: $( 3 , - 1 , - 3 ) \quad$ Vector: $\mathbf { v } = \{ 6,5,4 \}$
A) $6 x - 5 y + 4 z - 1 = 0$
B) $6 x + 5 y - 4 z + 1 = 0$
C) $6 x + 5 y + 4 z - 1 = 0$
D) $6 x + 5 y + 4 z + 1 = 0$
E) $6 x + 5 y + 4 z - 3 = 0$

Question

Quizplus · Accepted Answer

The equation of a plane in vector form is $ax + by + cz + d = 0$, where $(a, b, c)$ is a normal vector to the plane. Given the normal vector $\mathbf{v} = \{6, 5, 4\}$ and a point $(3, -1, -3)$, we substitute these into the plane equation to find $d$. Substituting $x = 3$, $y = -1$, and $z = -3$ into $6x + 5y + 4z + d = 0$, we get $18 - 5 - 12 + d = 0$, which simplifies to $d = -1$. Therefore, the correct equation is $6x + 5y + 4z - 1 = 0$.

Find an Equation of a Plane Passing Through the Given (3,−1,−3)( 3 , - 1 , - 3 ) \quad(3,−1,−3)

Find an Equation of a Plane Passing Through the Given $( 3 , - 1 , - 3 ) \quad$