Question

What is the equation for the plane illustrated below?

Answer 1

It is A 3x+3y+2z=6

Explanation:

Answer 2

Hence, none of the options presented are valid. The plane is represented by
`3 \cdot x + 3\cdot y + 2\cdot z = 6`.

Explanation:

The general equation in rectangular form for a 3-dimension plane is represented by:

`a\cdot x + b\cdot y + c\cdot z = d`

Where:

`x`,
`y`,
`z` - Orthogonal inputs.

`a`,
`b`,
`c`,
`d` - Plane constants.

The plane presented in the figure contains the following three points: (2, 0, 0), (0, 2, 0), (0, 0, 3)

For the determination of the resultant equation, three equations of line in three distinct planes orthogonal to each other. That is, expressions for the xy, yz and xz-planes with the resource of the general equation of the line:

xy-plane (2, 0, 0) and (0, 2, 0)

`y = m\cdot x + b`

`m = (y_(2)-y_(1))/(x_(2)-x_(1))`

Where:

`m` - Slope, dimensionless.

`x_(1)`,
`x_(2)` - Initial and final values for the independent variable, dimensionless.

`y_(1)`,
`y_(2)` - Initial and final values for the dependent variable, dimensionless.

`b` - x-Intercept, dimensionless.

If
`x_(1) = 2`,
`y_(1) = 0`,
`x_(2) = 0` and
`y_(2) = 2`, then:

Slope

`m = (2-0)/(0-2)`

`m = -1`

x-Intercept

`b = y_(1) - m\cdot x_(1)`

`b = 0 -(-1)\cdot (2)`

`b = 2`

The equation of the line in the xy-plane is
`y = -x+2` or
`x + y = 2`, which is equivalent to
`3\cdot x + 3\cdot y = 6`.

yz-plane (0, 2, 0) and (0, 0, 3)

`z = m\cdot y + b`

`m = (z_(2)-z_(1))/(y_(2)-y_(1))`

Where:

`m` - Slope, dimensionless.

`y_(1)`,
`y_(2)` - Initial and final values for the independent variable, dimensionless.

`z_(1)`,
`z_(2)` - Initial and final values for the dependent variable, dimensionless.

`b` - y-Intercept, dimensionless.

If
`y_(1) = 2`,
`z_(1) = 0`,
`y_(2) = 0` and
`z_(2) = 3`, then:

Slope

`m = (3-0)/(0-2)`

`m = -(3)/(2)`

y-Intercept

`b = z_(1) - m\cdot y_(1)`

`b = 0 -\left(-(3)/(2) \right)\cdot (2)`

`b = 3`

The equation of the line in the yz-plane is
`z = -(3)/(2)\cdot y+3` or
`3\cdot y + 2\cdot z = 6`.

xz-plane (2, 0, 0) and (0, 0, 3)

`z = m\cdot x + b`

`m = (z_(2)-z_(1))/(x_(2)-x_(1))`

Where:

`m` - Slope, dimensionless.

`x_(1)`,
`x_(2)` - Initial and final values for the independent variable, dimensionless.

`z_(1)`,
`z_(2)` - Initial and final values for the dependent variable, dimensionless.

`b` - z-Intercept, dimensionless.

If
`x_(1) = 2`,
`z_(1) = 0`,
`x_(2) = 0` and
`z_(2) = 3`, then:

Slope

`m = (3-0)/(0-2)`

`m = -(3)/(2)`

x-Intercept

`b = z_(1) - m\cdot x_(1)`

`b = 0 -\left(-(3)/(2) \right)\cdot (2)`

`b = 3`

The equation of the line in the xz-plane is
`z = -(3)/(2)\cdot x+3` or
`3\cdot x + 2\cdot z = 6`

After comparing each equation of the line to the definition of the equation of the plane, the following coefficients are obtained:

`a = 3`,
`b = 3`,
`c = 2`,
`d = 6`

Hence, none of the options presented are valid. The plane is represented by
`3 \cdot x + 3\cdot y + 2\cdot z = 6`.