The distance of the point `(2,1,-1)` from the plane `x-2y + 4z =9` is

Options

A. `( sqrt(13))/(21)`
B. `(13)/(21)`
C. `(13)/( sqrt( 21))`
D. `sqrt((13)/(21))`

Answer

To find the distance of the point \( (2, 1, -1) \) from the plane given by the equation \( x - 2y + 4z = 9 \), we can use the formula for the distance \( d \) from a point \( (x_0, y_0, z_0) \) to the plane \( Ax + By + Cz + D = 0 \):

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

### Step 1: Identify the coefficients from the plane equation
The plane equation can be rearranged to the standard form \( Ax + By + Cz + D = 0 \):
\[ x - 2y + 4z - 9 = 0 \]
From this, we can identify:
- \( A = 1 \)
- \( B = -2 \)
- \( C = 4 \)
- \( D = -9 \)

### Step 2: Substitute the point coordinates into the distance formula
The coordinates of the point are \( (x_0, y_0, z_0) = (2, 1, -1) \). Now we substitute these values into the distance formula:
\[ d = \frac{|1(2) + (-2)(1) + 4(-1) - 9|}{\sqrt{1^2 + (-2)^2 + 4^2}} \]

### Step 3: Calculate the numerator
Calculating the numerator:
\[ = |2 - 2 - 4 - 9| = |-13| = 13 \]

### Step 4: Calculate the denominator
Calculating the denominator:
\[ \sqrt{1^2 + (-2)^2 + 4^2} = \sqrt{1 + 4 + 16} = \sqrt{21} \]

### Step 5: Combine the results to find the distance
Now we can substitute the values of the numerator and denominator back into the distance formula:
\[ d = \frac{13}{\sqrt{21}} \]

### Final Answer
Thus, the distance of the point \( (2, 1, -1) \) from the plane \( x - 2y + 4z = 9 \) is:
\[ \frac{13}{\sqrt{21}} \]