Distance between plane 3x + 4y - 20 = 0 and point (0,0,-7) is :

Options

A. 4 units
B. 3 units
C. 2 units
D. 1 unit

Answer

To find the distance between the plane given by the equation \(3x + 4y - 20 = 0\) and the point \((0, 0, -7)\), 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 from the plane equation:**
The equation of the plane is \(3x + 4y - 20 = 0\). We can rearrange it to the standard form \(Ax + By + Cz + D = 0\):
- Here, \(A = 3\), \(B = 4\), \(C = 0\) (since there is no \(z\) term), and \(D = -20\).

2. **Substitute the coordinates of the point into the formula:**
The point we have is \((0, 0, -7)\), where \(x_0 = 0\), \(y_0 = 0\), and \(z_0 = -7\).
- Substitute these values into the distance formula:
\[ D = \frac{|3(0) + 4(0) + 0(-7) - 20|}{\sqrt{3^2 + 4^2 + 0^2}} \]

3. **Calculate the numerator:**
- The numerator simplifies to:
\[ |0 + 0 + 0 - 20| = |-20| = 20 \]

4. **Calculate the denominator:**
- The denominator simplifies to:
\[ \sqrt{3^2 + 4^2 + 0^2} = \sqrt{9 + 16 + 0} = \sqrt{25} = 5 \]

5. **Calculate the distance:**
- Now, substituting the values into the distance formula:
\[ D = \frac{20}{5} = 4 \]

Thus, the distance between the plane \(3x + 4y - 20 = 0\) and the point \((0, 0, -7)\) is \(4\) units.

### Final Answer:
The distance is \(4\) units.