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 1: $(6, -1, 4) - (5, -5, -6) = (1, 4, 10)$Vector 2: $(-2, -5, 5) - (5, -5, -6) = (-7, 0, 11)$Next, we find the normal vector to the plane by taking the cross product of these two vectors:$\vec{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 4 & 10 \ -7 & 0 & 11 \end{vmatrix} = (44\mathbf{i} - 81\mathbf{j} + 28\mathbf{k})$This gives us the normal vector $(44, -81, 28)$, which are the coefficients of $x$, $y$, and $z$ in the plane equation. To find the constant term, we use one of the points through which the plane passes, say $(5, -5, -6)$:$44(5) - 81(-5) + 28(-6) = 220 + 405 - 168 = 457$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 $( 5 , - 5 , - 6 ) , ( 6 , - 1,4 ) , ( - 2 , - 5,5 )$