Find 'k' so that the distance between the points (7,1,-3) and (4,5,k) be 13 units

Answer

To find the value of \( k \) such that the distance between the points \( (7, 1, -3) \) and \( (4, 5, k) \) is 13 units, we can follow these steps:

### Step 1: Use the Distance Formula
The distance \( d \) between two points \( (x_1, y_1, z_1) \) and \( (x_2, y_2, z_2) \) 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 have:
- \( (x_1, y_1, z_1) = (7, 1, -3) \)
- \( (x_2, y_2, z_2) = (4, 5, k) \)

### Step 2: Set Up the Equation
Substituting the coordinates into the distance formula, we have:
\[ \sqrt{(4 - 7)^2 + (5 - 1)^2 + (k - (-3))^2} = 13 \]
This simplifies to:
\[ \sqrt{(-3)^2 + (4)^2 + (k + 3)^2} = 13 \]

### Step 3: Square Both Sides
To eliminate the square root, we square both sides:
\[ (-3)^2 + (4)^2 + (k + 3)^2 = 13^2 \]
Calculating the squares gives:
\[ 9 + 16 + (k + 3)^2 = 169 \]
This simplifies to:
\[ 25 + (k + 3)^2 = 169 \]

### Step 4: Isolate the Quadratic Term
Now, we isolate \( (k + 3)^2 \):
\[ (k + 3)^2 = 169 - 25 \]
Calculating the right side gives:
\[ (k + 3)^2 = 144 \]

### Step 5: Solve for \( k + 3 \)
Taking the square root of both sides, we have:
\[ k + 3 = \pm 12 \]

### Step 6: Solve for \( k \)
Now we solve for \( k \):
1. \( k + 3 = 12 \) leads to \( k = 12 - 3 = 9 \)
2. \( k + 3 = -12 \) leads to \( k = -12 - 3 = -15 \)

### Final Values of \( k \)
Thus, the two possible values of \( k \) are:
\[ k = 9 \quad \text{and} \quad k = -15 \]