A point on y-axis at a distance `sqrt(20)` units from the
point (2, 2, 5) is

Options

A. (1, 1, 2)
B. (0, 4, 0)
C. (4, -4, 0)
D. None of these

Answer

To find the point on the y-axis that is at a distance of \(\sqrt{20}\) units from the point \(A(2, 2, 5)\), we can follow these steps:

### Step 1: Define the point on the y-axis
A point on the y-axis can be represented as \(P(0, y, 0)\), where \(y\) is the y-coordinate we need to find.

### Step 2: Use the distance formula
The distance \(d\) between two points \(A(x_1, y_1, z_1)\) and \(B(x_2, y_2, z_2)\) in three-dimensional space is given by the formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]
In our case, we need to find the distance between point \(A(2, 2, 5)\) and point \(P(0, y, 0)\).

### Step 3: Set up the distance equation
Substituting the coordinates into the distance formula, we have:
\[ \sqrt{(0 - 2)^2 + (y - 2)^2 + (0 - 5)^2} = \sqrt{20} \]
This simplifies to:
\[ \sqrt{(-2)^2 + (y - 2)^2 + (-5)^2} = \sqrt{20} \]
\[ \sqrt{4 + (y - 2)^2 + 25} = \sqrt{20} \]

### Step 4: Simplify the equation
Now, simplify the left side:
\[ \sqrt{29 + (y - 2)^2} = \sqrt{20} \]

### Step 5: Square both sides
To eliminate the square root, we square both sides:
\[ 29 + (y - 2)^2 = 20 \]

### Step 6: Rearrange the equation
Now, rearranging the equation gives:
\[ (y - 2)^2 = 20 - 29 \]
\[ (y - 2)^2 = -9 \]

### Step 7: Solve for \(y\)
Since the square of a real number cannot be negative, we find that:
\[ y - 2 = \pm \sqrt{-9} \]
This implies:
\[ y - 2 = \pm 3i \]
Thus, we have:
\[ y = 2 \pm 3i \]

### Conclusion
The point on the y-axis that is at a distance of \(\sqrt{20}\) units from the point \(A(2, 2, 5)\) does not yield a real solution, indicating that there are no real points on the y-axis at that distance from \(A\).