Question

2. The general equation of a plane is Ax + By + Cz = D, where A, B, C, and D are real numbers and A is nonnegative. Find the equation of the plane containing the points (3, 0, 0), (0, 8, 0), and (0, 0, 6). Show each step of your process. Then graph the plane.

Answer 1

z=6-(3/4)y-2x
z+(3/4)y+2x=6

By connecting the three points in the graph, Got this equation by isolating each plane to figure it out.

Answer 2

The standard equation is 8x+3y+4z= -24.

Explanation:

Given : The general equation of a plane is Ax + By + Cz = D, where A, B, C, and D are real numbers and A is non negative.

We have to find the equation of the plane containing the points (3, 0, 0), (0, 8, 0), and (0, 0, 6).

If given coordinates of three points
`A(x_1, y_1, z_1), B(x_2, y_2, z_2),C(x_3, y_3, z_3)` lying on a plane are defined then the plane equation can be found using the formula,

`\left[\begin{array}{ccc}x - x_1&y-y_1&z-z_1\\x_2 - x_1&y_2-y_1&z_2-z_1\\x_3 - x_1&y_3-y_1&z_3-z_1\end{array}\right]=0`

For the given coordinates, A(3, 0, 0), B(0, 8, 0), and C(0, 0, 6)

Substitute, we have,

`\begin{pmatrix}x-3&y-0&z-0\\ \:0-3&8-0&0-0\\ \:0-3&0-0&6-0\end{pmatrix}=0`

Solving further, we get,

`\begin{pmatrix}x-3&y&z\\ \:-3&8&0\\ \:-3&0&6\end{pmatrix}=0`

Evaluate along row !, we get,

`(x-3)(8\cdot6-0) -y((-3)\cdot6-0)+z(0-8\cdot (-3))`

We get,

`48(x-3)+18y+24z=0`

Simplify , we get,

`8x+3y+4z -24 = 0`

Thus, the standard equation is 8x+3y+4z= -24