The distance of a plane `3x + 4y - 5z = 60 ` from
origin

Options

A. `3/sqrt(50)`
B. `6/sqrt(50)`
C. `6sqrt(2)`
D. `4/(5sqrt(2))`

Answer

To find the distance of the plane given by the equation \(3x + 4y - 5z = 60\) from the origin \((0, 0, 0)\), we can use the formula for the distance \(D\) from a point to a plane. The formula is:

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

where \(ax + by + cz = d\) is the equation of the plane, and \((x_1, y_1, z_1)\) is the point from which we want to find the distance.

### Step-by-Step Solution:

1. **Identify the coefficients from the plane equation:**
The equation of the plane is \(3x + 4y - 5z = 60\). Here, we have:
- \(a = 3\)
- \(b = 4\)
- \(c = -5\)
- \(d = 60\)

2. **Substitute the coordinates of the origin into the formula:**
The coordinates of the origin are \((x_1, y_1, z_1) = (0, 0, 0)\). Substitute these values into the distance formula:
\[ D = \frac{|3(0) + 4(0) - 5(0) - 60|}{\sqrt{3^2 + 4^2 + (-5)^2}} \]

3. **Simplify the numerator:**
The numerator simplifies as follows:
\[ D = \frac{|0 + 0 + 0 - 60|}{\sqrt{3^2 + 4^2 + 25}} = \frac{|-60|}{\sqrt{9 + 16 + 25}} = \frac{60}{\sqrt{50}} \]

4. **Calculate the denominator:**
Now calculate \(\sqrt{50}\):
\[ \sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2} \]

5. **Combine the results:**
Substitute back into the distance formula:
\[ D = \frac{60}{5\sqrt{2}} = \frac{12}{\sqrt{2}} = 12 \cdot \frac{\sqrt{2}}{2} = 6\sqrt{2} \]

### Final Answer:
The distance of the plane \(3x + 4y - 5z = 60\) from the origin is \(6\sqrt{2}\).

---