What is the distance of the origin from the plane `2x+6y-3z+7=0`?

Options

A. 1
B. 2
C. 3
D. 6

Answer

To find the distance of the origin from the plane given by the equation \(2x + 6y - 3z + 7 = 0\), we can use the formula for the distance \(d\) from a point \((x_1, y_1, z_1)\) to the plane \(ax + by + cz + d = 0\):

\[ d = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}} \]

### Step-by-Step Solution:

1. **Identify the coefficients of the plane equation**:
The given plane equation is \(2x + 6y - 3z + 7 = 0\). Here, we can identify:
- \(a = 2\)
- \(b = 6\)
- \(c = -3\)
- \(d = 7\)

2. **Identify the coordinates of the point**:
The point we are interested in is the origin, which has coordinates \((x_1, y_1, z_1) = (0, 0, 0)\).

3. **Substitute the point's coordinates into the distance formula**:
Substitute \(x_1 = 0\), \(y_1 = 0\), \(z_1 = 0\) into the formula:
\[ d = \frac{|2(0) + 6(0) - 3(0) + 7|}{\sqrt{2^2 + 6^2 + (-3)^2}} \]

4. **Calculate the numerator**:
The numerator simplifies to:
\[ |0 + 0 + 0 + 7| = |7| = 7 \]

5. **Calculate the denominator**:
The denominator simplifies to:
\[ \sqrt{2^2 + 6^2 + (-3)^2} = \sqrt{4 + 36 + 9} = \sqrt{49} = 7 \]

6. **Calculate the distance**:
Now, substitute the values back into the distance formula:
\[ d = \frac{7}{7} = 1 \]

### Final Answer:
The distance of the origin from the plane \(2x + 6y - 3z + 7 = 0\) is \(1\).

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