Find the general form of the equation of the plane passing through the three points. [Be sure to reduce the coefficients in your answer to lowest terms by dividing out any common factor.]
$$( 5,2 , - 1 ) , ( 3,3,2 ) , ( 4,4,1 )$$
A) $4 x - y + 3 z - 15 = 0$
B) $4 x - y + 3 z = 0$
C) $5 x + 2 y - z = 0$
D) $4 x - y + 3 z + 15 = 0$
E) $5 x + 2 y - z - 15 = 0$

Question

Quizplus · Accepted Answer

To find the equation of the plane passing through the points $(5,2,-1)$, $(3,3,2)$, and $(4,4,1)$, we first find two vectors that lie in the plane by subtracting coordinates of the points: $\vec{AB} = (3-5, 3-2, 2+1) = (-2, 1, 3)$ and $\vec{AC} = (4-5, 4-2, 1+1) = (-1, 2, 2)$. The normal vector to the plane, $\vec{n}$, can be found by taking the cross product of $\vec{AB}$ and $\vec{AC}$: $\vec{n} = \vec{AB} \times \vec{AC} = (1(2)-3(2), -((-2)(2)-3(-1)), (-2)(2)-1(1)) = (-4, -1, 3)$. The equation of the plane is given by $\vec{n} \cdot (\vec{r} - \vec{r_0}) = 0$, where $\vec{r}$ is a general point $(x, y, z)$ on the plane, and $\vec{r_0}$ is a specific point on the plane (we can use $(5,2,-1)$). Plugging in the values, we get $(-4, -1, 3) \cdot ((x-5), (y-2), (z+1)) = 0$, which simplifies to $-4(x-5) - (y-2) + 3(z+1) = 0$, or $4x - y + 3z - 15 = 0$ after distributing and rearranging.

Find the General Form of the Equation of the Plane (5,2,−1),(3,3,2),(4,4,1)( 5,2 , - 1 ) , ( 3,3,2 ) , ( 4,4,1 )(5,2,−1),(3,3,2),(4,4,1)

Find the General Form of the Equation of the Plane $( 5,2 , - 1 ) , ( 3,3,2 ) , ( 4,4,1 )$