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.]
$$( 5 , - 5 , - 6 ) , ( 6 , - 1,4 ) , ( - 2 , - 5,5 )$$
A) $44 x - 81 y + 28 z - 457 = 0$
B) $44 x - 81 y + 28 z = 0$
C) $5 x - 5 y - 6 z = 0$
D) $44 x - 81 y + 28 z + 457 = 0$
E) $5 x - 5 y - 6 z + 457 = 0$

Question

Quizplus · Accepted Answer

To find the equation of the plane passing through the points $(5, -5, -6)$, $(6, -1, 4)$, and $(-2, -5, 5)$, we first find two vectors that lie in the plane by subtracting the coordinates of the points:Vector $\mathbf{AB} = (6-5, -1+5, 4+6) = (1, 4, 10)$Vector $\mathbf{AC} = (-2-5, -5+5, 5+6) = (-7, 0, 11)$Next, we find the normal vector to the plane by taking the cross product of $\mathbf{AB}$ and $\mathbf{AC}$:$\mathbf{N} = \mathbf{AB} \times \mathbf{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 4 & 10 \ -7 & 0 & 11 \end{vmatrix} = (4*11 - 10*0)\mathbf{i} - (1*11 - 10*(-7))\mathbf{j} + (1*0 - 4*(-7))\mathbf{k} = (44)\mathbf{i} - (81)\mathbf{j} + (28)\mathbf{k}$So, the normal vector is $(44, -81, 28)$, and the general form of the plane equation is $44x - 81y + 28z + D = 0$. To find $D$, we substitute one of the points into the equation. Using $(5, -5, -6)$:$44(5) - 81(-5) + 28(-6) + D = 0$$220 + 405 - 168 + D = 0$$457 + D = 0$$D = -457$Thus, the equation of the plane is $44x - 81y + 28z - 457 = 0$.

Find the General Form of the Equation of the Plane (5,−5,−6),(6,−1,4),(−2,−5,5)( 5 , - 5 , - 6 ) , ( 6 , - 1,4 ) , ( - 2 , - 5,5 )(5,−5,−6),(6,−1,4),(−2,−5,5)

Find the General Form of the Equation of the Plane $( 5 , - 5 , - 6 ) , ( 6 , - 1,4 ) , ( - 2 , - 5,5 )$