Find an equation of a plane passing through the following three points. $$( 1,2 , - 2 ) , ( - 4,2,5 ) , ( 3,4,5 )$$
A) $14 x + 49 y + 10 z + 104 = 0$
B) $14 x - 49 y + 10 z - 104 = 0$
C) $- 14 x - 49 y + 10 z + 104 = 0$
D) $14 x - 49 y - 10 z + 104 = 0$
E) $14 x - 49 y + 10 z + 104 = 0$

Question

Find  an equation of a plane passing through the following three points. $$( 1,2 , - 2 ) , ( - 4,2,5 ) , ( 3,4,5 )$$
A) $14 x + 49 y + 10 z + 104 = 0$
B) $14 x - 49 y + 10 z - 104 = 0$
C) $- 14 x - 49 y + 10 z + 104 = 0$
D) $14 x - 49 y - 10 z + 104 = 0$
E) $14 x - 49 y + 10 z + 104 = 0$

Quizplus · Accepted Answer

To find the equation of a plane passing through three points $P_1(1,2,-2)$, $P_2(-4,2,5)$, and $P_3(3,4,5)$, we first find two vectors that lie in the plane by subtracting the coordinates of these points: $\vec{P_1P_2} = P_2 - P_1 = (-4-1, 2-2, 5+2) = (-5, 0, 7)$ and $\vec{P_1P_3} = P_3 - P_1 = (3-1, 4-2, 5+2) = (2, 2, 7)$. Next, we find the normal vector $\vec{n}$ to the plane by taking the cross product of $\vec{P_1P_2}$ and $\vec{P_1P_3}$: $\vec{n} = \vec{P_1P_2} \times \vec{P_1P_3} = \begin{vmatrix} i & j & k \ -5 & 0 & 7 \ 2 & 2 & 7 \end{vmatrix} = (0*7 - 7*2)i - (-5*7 - 7*2)j + (-5*2 - 0*2)k = (-14)i + 49j - 10k$. Therefore, the normal vector is $\vec{n} = (-14, 49, -10)$. The equation of a plane is given by $n_x(x-x_0) + n_y(y-y_0) + n_z(z-z_0) = 0$, where $(x_0, y_0, z_0)$ is a point on the plane (we can use any of the given points, for example, $P_1(1,2,-2)$) and $(n_x, n_y, n_z)$ are the components of the normal vector. Substituting the values, we get $-14(x-1) + 49(y-2) - 10(z+2) = 0$, which simplifies to $-14x + 14 + 49y - 98 - 10z - 20 = 0$, or $14x - 49y + 10z + 104 = 0$, matching option E.

Find an Equation of a Plane Passing Through the Following