The points on z-axis which are at a distance of 6 units from the point (-4,2,-1) are

Options

A. `(0,0, pm 5)`
B. `(0,0,pm 3)`
C. (0,0,3), (0,0,-5)
D. (0,0,-3), (0,0,5)

Answer

To find the points on the z-axis that are at a distance of 6 units from the point (-4, 2, -1), we can follow these steps:

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

### Step 2: Use the distance formula
The distance \( d \) between the point \( P(0, 0, a) \) and the point \( A(-4, 2, -1) \) can be calculated using the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]
Substituting the coordinates:
\[ d = \sqrt{(0 - (-4))^2 + (0 - 2)^2 + (a - (-1))^2} \]
This simplifies to:
\[ d = \sqrt{(4)^2 + (-2)^2 + (a + 1)^2} \]
\[ d = \sqrt{16 + 4 + (a + 1)^2} \]
\[ d = \sqrt{20 + (a + 1)^2} \]

### Step 3: Set the distance equal to 6
Since we know the distance \( d \) is 6 units, we set up the equation:
\[ \sqrt{20 + (a + 1)^2} = 6 \]

### Step 4: Square both sides
To eliminate the square root, we square both sides:
\[ 20 + (a + 1)^2 = 36 \]

### Step 5: Solve for \( (a + 1)^2 \)
Subtract 20 from both sides:
\[ (a + 1)^2 = 36 - 20 \]
\[ (a + 1)^2 = 16 \]

### Step 6: Take the square root
Taking the square root of both sides gives:
\[ a + 1 = \pm 4 \]

### Step 7: Solve for \( a \)
Now we solve for \( a \):
1. If \( a + 1 = 4 \):
\[ a = 4 - 1 = 3 \]
2. If \( a + 1 = -4 \):
\[ a = -4 - 1 = -5 \]

### Step 8: Write the points on the z-axis
Thus, the points on the z-axis that are 6 units away from the point (-4, 2, -1) are:
\[ (0, 0, 3) \quad \text{and} \quad (0, 0, -5) \]

### Final Answer
The points on the z-axis are \( (0, 0, 3) \) and \( (0, 0, -5) \).

---