The distance of point `(2,-1,0) from the plane `2x+y+2z+8=0` is

Options

A. `(7)/(5)`
B. `(5)/(3)`
C. `(8)/(3)`
D. `(11)/(3)`

Answer

To find the distance of the point \( P(2, -1, 0) \) from the plane given by the equation \( 2x + y + 2z + 8 = 0 \), 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 1: Identify the coefficients from the plane equation
From the plane equation \( 2x + y + 2z + 8 = 0 \), we can identify:
- \( A = 2 \)
- \( B = 1 \)
- \( C = 2 \)
- \( D = 8 \)

### Step 2: Substitute the point coordinates into the formula
The coordinates of the point \( P \) are \( (x_0, y_0, z_0) = (2, -1, 0) \). We substitute these values into the formula:

\[ d = \frac{|2(2) + 1(-1) + 2(0) + 8|}{\sqrt{2^2 + 1^2 + 2^2}} \]

### Step 3: Calculate the numerator
Calculating the numerator:

\[ 2(2) + 1(-1) + 2(0) + 8 = 4 - 1 + 0 + 8 = 11 \]

So, the absolute value is:

\[ |11| = 11 \]

### Step 4: Calculate the denominator
Calculating the denominator:

\[ \sqrt{2^2 + 1^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \]

### Step 5: Calculate the distance
Now substituting back into the distance formula:

\[ d = \frac{11}{3} \]

Thus, the distance of the point \( (2, -1, 0) \) from the plane \( 2x + y + 2z + 8 = 0 \) is:

\[ \boxed{\frac{11}{3}} \]