Question

CAN anyone please help a girl out with this question?????? I really am having a realllllllllyyyy hard time and Itd be amazing if someone could help.

The general equation of a plane is Ax+By+Cz=D. A B C D are real numbers, A is nonnegative. Find the equation of the plane containing (3,0,0) (0,8,0) and (0,0,6) Show your steps, then graph.

Answer 1

first find a vector between two points, then find a vector between one of those two points and the remaining point, then find the cross product to get normal vector

Answer 2

Try plug in points.

\[(3,0,0)\quad\Longrightarrow \quad A(3)+B(0)+C(0) = D\quad\Rightarrow\quad3A = D\\
(0,8,0)\quad\Longrightarrow \quad A(0)+B(8)+C(0) = D\quad\Rightarrow\quad8B = D\\
(0,0,6)\quad\Longrightarrow \quad A(0)+B(0)+C(6) = D\quad\Rightarrow\quad6C = D\\\]

So you have \(3A = 8B = 6C = D\).

Hmm. A could be \(\dfrac{1}{3}\), B could be \(\dfrac{1}{8}\), C could be \(\dfrac{1}{6}, \) and \(D = 1\)

So you have \(\dfrac{1}{3}x+\dfrac{1}{8}y+\dfrac{1}{6}z = 1\)

Multiply both sides by \(LCM(3, 8, 6) = 24\) to have a better looking equation.

\(\Large\boxed{8x + 3y + 4z = 24}\)