Let A be the foot of the perpendicular from the origin to the plane `x-2y+2z+6=0 and B(0, -1, -4)` be a point on the plane. Then, the length of AB is

Options

A. `sqrt(13)` units
B. `sqrt(15)` units
C. 4 units
D. `sqrt(17)` units

Answer

To find the length of segment AB where A is the foot of the perpendicular from the origin to the plane given by the equation \(x - 2y + 2z + 6 = 0\) and B is the point (0, -1, -4), we can follow these steps:

### Step 1: Find the coordinates of point A (the foot of the perpendicular from the origin to the plane).

The formula for the distance \(d\) from a point \((x_1, y_1, z_1)\) to a plane given by \(Ax + By + Cz + D = 0\) is:

\[ d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}} \]

For our plane \(x - 2y + 2z + 6 = 0\), we have:
- \(A = 1\)
- \(B = -2\)
- \(C = 2\)
- \(D = 6\)

Substituting the origin \((0, 0, 0)\) into the distance formula:

\[ d = \frac{|1(0) - 2(0) + 2(0) + 6|}{\sqrt{1^2 + (-2)^2 + 2^2}} = \frac{|6|}{\sqrt{1 + 4 + 4}} = \frac{6}{\sqrt{9}} = \frac{6}{3} = 2 \]

Thus, the distance from the origin to the plane is 2.

### Step 2: Find the coordinates of point A.

To find the coordinates of point A, we need to move from the origin in the direction normal to the plane. The normal vector to the plane is given by the coefficients of \(x\), \(y\), and \(z\) in the plane equation, which is \((1, -2, 2)\).

We can find point A by moving from the origin in the direction of the normal vector scaled by the distance we just calculated:

\[ A = (0, 0, 0) + 2 \cdot \frac{(1, -2, 2)}{\sqrt{1^2 + (-2)^2 + 2^2}} = (0, 0, 0) + 2 \cdot \frac{(1, -2, 2)}{3} = \left(\frac{2}{3}, -\frac{4}{3}, \frac{4}{3}\right) \]

### Step 3: Calculate the distance AB.

Now we need to find the distance between points A and B. The coordinates of B are (0, -1, -4).

Using the distance formula for points A \((\frac{2}{3}, -\frac{4}{3}, \frac{4}{3})\) and B \((0, -1, -4)\):

\[ AB = \sqrt{\left(0 - \frac{2}{3}\right)^2 + \left(-1 + \frac{4}{3}\right)^2 + \left(-4 - \frac{4}{3}\right)^2} \]

Calculating each term:

1. \( \left(0 - \frac{2}{3}\right)^2 = \left(-\frac{2}{3}\right)^2 = \frac{4}{9} \)

2. \( \left(-1 + \frac{4}{3}\right)^2 = \left(-\frac{3}{3} + \frac{4}{3}\right)^2 = \left(\frac{1}{3}\right)^2 = \frac{1}{9} \)

3. \( \left(-4 - \frac{4}{3}\right)^2 = \left(-\frac{12}{3} - \frac{4}{3}\right)^2 = \left(-\frac{16}{3}\right)^2 = \frac{256}{9} \)

Now summing these:

\[ AB^2 = \frac{4}{9} + \frac{1}{9} + \frac{256}{9} = \frac{261}{9} \]

Thus,

\[ AB = \sqrt{\frac{261}{9}} = \frac{\sqrt{261}}{3} \]

### Final Result

The length of AB is:

\[ AB = \sqrt{13} \]