Distance between the point (0,1,7) and the plane 3x + 4y + 1 = 0 is :

Options

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

Answer

To find the distance between the point \( P(0, 1, 7) \) and the plane given by the equation \( 3x + 4y + 1 = 0 \), we can use the formula for the distance \( d \) from a point \( (x_1, y_1, z_1) \) to a plane given by the equation \( Ax + By + Cz + D = 0 \):

\[ d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}} \]

### Step-by-Step Solution:

1. **Identify the coefficients from the plane equation**:
The equation of the plane is \( 3x + 4y + 1 = 0 \). Here, we can identify:
- \( A = 3 \)
- \( B = 4 \)
- \( C = 0 \) (since there is no \( z \) term)
- \( D = 1 \)

2. **Substitute the coordinates of the point**:
The coordinates of the point \( P \) are \( (x_1, y_1, z_1) = (0, 1, 7) \).

3. **Calculate \( Ax_1 + By_1 + Cz_1 + D \)**:
\[ Ax_1 + By_1 + Cz_1 + D = 3(0) + 4(1) + 0(7) + 1 = 0 + 4 + 0 + 1 = 5 \]

4. **Calculate the denominator \( \sqrt{A^2 + B^2 + C^2} \)**:
\[ \sqrt{A^2 + B^2 + C^2} = \sqrt{3^2 + 4^2 + 0^2} = \sqrt{9 + 16 + 0} = \sqrt{25} = 5 \]

5. **Substitute into the distance formula**:
\[ d = \frac{|5|}{5} = \frac{5}{5} = 1 \]

### Final Answer:
The distance between the point \( (0, 1, 7) \) and the plane \( 3x + 4y + 1 = 0 \) is \( 1 \) unit.

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