Find the distance of the plane `2x - 2y + 4z = 6` from the origin.

Options

A. `(8)/(sqrt(29))`
B. `(6)/(sqrt(29))`
C. `(4)/(sqrt(29))`
D. `(2)/(sqrt(29))`

Answer

To find the distance of the plane \(2x - 2y + 4z = 6\) from the origin, we can use the formula for the distance \(D\) from a point \((x_0, y_0, z_0)\) to a plane given by the equation \(Ax + By + Cz + D = 0\):

\[ D = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}} \]

### Step-by-Step Solution:

1. **Identify the coefficients of the plane equation**:
The given plane equation is \(2x - 2y + 4z = 6\). We can rewrite this in the standard form \(Ax + By + Cz + D = 0\):
\[ 2x - 2y + 4z - 6 = 0 \]
Here, \(A = 2\), \(B = -2\), \(C = 4\), and \(D = -6\).

2. **Identify the coordinates of the point**:
We need to find the distance from the origin, which has coordinates \((x_0, y_0, z_0) = (0, 0, 0)\).

3. **Substitute the values into the distance formula**:
Now, substituting \(x_0 = 0\), \(y_0 = 0\), \(z_0 = 0\) into the distance formula:
\[ D = \frac{|2(0) - 2(0) + 4(0) - 6|}{\sqrt{2^2 + (-2)^2 + 4^2}} \]

4. **Calculate the numerator**:
The numerator becomes:
\[ |0 - 0 + 0 - 6| = |-6| = 6 \]

5. **Calculate the denominator**:
The denominator is:
\[ \sqrt{2^2 + (-2)^2 + 4^2} = \sqrt{4 + 4 + 16} = \sqrt{24} = 2\sqrt{6} \]

6. **Combine to find the distance**:
Now we can find the distance \(D\):
\[ D = \frac{6}{2\sqrt{6}} = \frac{3}{\sqrt{6}} = \frac{3\sqrt{6}}{6} = \frac{\sqrt{6}}{2} \]

### Final Answer:
The distance of the plane \(2x - 2y + 4z = 6\) from the origin is \(\frac{\sqrt{6}}{2}\).