If the distance between the points `(7,1,-3) and (4,5,lambda)` is 13 units. Then what is one of the values of `lambda`

Options

A. 20
B. 10
C. 9
D. 8

Answer

To solve the problem, we need to find the value of \( \lambda \) such that the distance between the points \( (7, 1, -3) \) and \( (4, 5, \lambda) \) is 13 units. We will use the distance formula in three-dimensional space.

### Step-by-Step Solution:

1. **Identify the Points:**
Let \( P_1 = (7, 1, -3) \) and \( P_2 = (4, 5, \lambda) \).

2. **Write the Distance Formula:**
The distance \( d \) between two points \( (x_1, y_1, z_1) \) and \( (x_2, y_2, z_2) \) in 3D is given by:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]

3. **Substitute the Points into the Formula:**
Here, \( x_1 = 7, y_1 = 1, z_1 = -3 \) and \( x_2 = 4, y_2 = 5, z_2 = \lambda \).
\[ d = \sqrt{(4 - 7)^2 + (5 - 1)^2 + (\lambda - (-3))^2} \]

4. **Calculate the Differences:**
\[ d = \sqrt{(-3)^2 + (4)^2 + (\lambda + 3)^2} \]
\[ d = \sqrt{9 + 16 + (\lambda + 3)^2} \]
\[ d = \sqrt{25 + (\lambda + 3)^2} \]

5. **Set the Distance Equal to 13:**
Since the distance is given as 13 units:
\[ \sqrt{25 + (\lambda + 3)^2} = 13 \]

6. **Square Both Sides:**
To eliminate the square root, we square both sides:
\[ 25 + (\lambda + 3)^2 = 169 \]

7. **Isolate the Square Term:**
\[ (\lambda + 3)^2 = 169 - 25 \]
\[ (\lambda + 3)^2 = 144 \]

8. **Take the Square Root:**
Taking the square root of both sides gives:
\[ \lambda + 3 = \pm 12 \]

9. **Solve for \( \lambda \):**
This gives us two equations:
- \( \lambda + 3 = 12 \)
- \( \lambda + 3 = -12 \)

For the first equation:
\[ \lambda = 12 - 3 = 9 \]

For the second equation:
\[ \lambda = -12 - 3 = -15 \]

10. **Conclusion:**
Therefore, one of the values of \( \lambda \) is \( 9 \) and the other is \( -15 \).