What is the perpendicular distance of plane 2x-y+3z =10 from origin

Answer

To find the perpendicular distance of the plane given by the equation \(2x - y + 3z = 10\) from the origin, we can use the formula for the distance \(d\) from a point to a plane. The formula is given by:

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

where \(Ax + By + Cz + D = 0\) is the equation of the plane, and \((x_0, y_0, z_0)\) is the point from which we are measuring the distance (in this case, the origin \((0, 0, 0)\)).

### Step-by-Step Solution:

1. **Identify the coefficients from the plane equation:**
The given plane equation is \(2x - y + 3z = 10\). We can rewrite it in the form \(Ax + By + Cz + D = 0\):
\[ 2x - y + 3z - 10 = 0 \]
Here, \(A = 2\), \(B = -1\), \(C = 3\), and \(D = -10\).

2. **Substitute the coordinates of the origin:**
The coordinates of the origin are \((x_0, y_0, z_0) = (0, 0, 0)\).

3. **Calculate the numerator of the distance formula:**
Substitute \(x_0\), \(y_0\), and \(z_0\) into the formula:
\[ |Ax_0 + By_0 + Cz_0 + D| = |2(0) + (-1)(0) + 3(0) - 10| = |-10| = 10 \]

4. **Calculate the denominator of the distance formula:**
Now, calculate \(\sqrt{A^2 + B^2 + C^2}\):
\[ \sqrt{A^2 + B^2 + C^2} = \sqrt{2^2 + (-1)^2 + 3^2} = \sqrt{4 + 1 + 9} = \sqrt{14} \]

5. **Combine the results to find the distance:**
Now, substitute the values into the distance formula:
\[ d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}} = \frac{10}{\sqrt{14}} \]

### Final Answer:
The perpendicular distance of the plane \(2x - y + 3z = 10\) from the origin is:
\[ \frac{10}{\sqrt{14}} \text{ units} \]