Find the points on the line through the points A (1,2,3) and B ( 5,8 ,15) at a distance of 14 units from the mid-point of AB.

Answer

To find the points on the line through the points A(1, 2, 3) and B(5, 8, 15) that are at a distance of 14 units from the midpoint of AB, we can follow these steps:

### Step 1: Find the Midpoint of AB
The midpoint \( M \) of points A and B can be calculated using the midpoint formula:
\[ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right) \]
Substituting the coordinates of A(1, 2, 3) and B(5, 8, 15):
\[ M = \left( \frac{1 + 5}{2}, \frac{2 + 8}{2}, \frac{3 + 15}{2} \right) = \left( \frac{6}{2}, \frac{10}{2}, \frac{18}{2} \right) = (3, 5, 9) \]

### Step 2: Find the Direction Ratios of the Line AB
The direction ratios of the line through points A and B can be found by subtracting the coordinates of A from B:
\[ \text{Direction Ratios} = (5 - 1, 8 - 2, 15 - 3) = (4, 6, 12) \]

### Step 3: Parametric Equations of the Line
Using the direction ratios, we can write the parametric equations of the line:
\[ x = 1 + 4t, \quad y = 2 + 6t, \quad z = 3 + 12t \]
where \( t \) is a parameter.

### Step 4: Find Points at a Distance of 14 Units from M
We need to find points on the line that are at a distance of 14 units from the midpoint \( M(3, 5, 9) \). The distance formula in three-dimensional space is given by:
\[ d = \sqrt{(x - x_m)^2 + (y - y_m)^2 + (z - z_m)^2} \]
Setting \( d = 14 \) and substituting \( M(3, 5, 9) \):
\[ \sqrt{(1 + 4t - 3)^2 + (2 + 6t - 5)^2 + (3 + 12t - 9)^2} = 14 \]
Squaring both sides:
\[ (1 + 4t - 3)^2 + (2 + 6t - 5)^2 + (3 + 12t - 9)^2 = 196 \]
This simplifies to:
\[ (-2 + 4t)^2 + (-3 + 6t)^2 + (-6 + 12t)^2 = 196 \]

### Step 5: Expand and Simplify
Expanding each term:
\[ (4t - 2)^2 = 16t^2 - 16t + 4 \]
\[ (6t - 3)^2 = 36t^2 - 36t + 9 \]
\[ (12t - 6)^2 = 144t^2 - 144t + 36 \]
Combining these:
\[ 16t^2 - 16t + 4 + 36t^2 - 36t + 9 + 144t^2 - 144t + 36 = 196 \]
\[ 196t^2 - 196t + 49 = 196 \]
Subtracting 196 from both sides:
\[ 196t^2 - 196t - 147 = 0 \]

### Step 6: Solve the Quadratic Equation
Dividing the entire equation by 49:
\[ 4t^2 - 4t - 3 = 0 \]
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ t = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 4 \cdot (-3)}}{2 \cdot 4} = \frac{4 \pm \sqrt{16 + 48}}{8} = \frac{4 \pm \sqrt{64}}{8} = \frac{4 \pm 8}{8} \]
Calculating the two possible values for \( t \):
1. \( t = \frac{12}{8} = 1.5 \)
2. \( t = \frac{-4}{8} = -0.5 \)

### Step 7: Find the Corresponding Points
Substituting \( t = 1.5 \) into the parametric equations:
\[ x = 1 + 4(1.5) = 7, \quad y = 2 + 6(1.5) = 11, \quad z = 3 + 12(1.5) = 21 \]
So one point is \( (7, 11, 21) \).

Substituting \( t = -0.5 \):
\[ x = 1 + 4(-0.5) = -1, \quad y = 2 + 6(-0.5) = -1, \quad z = 3 + 12(-0.5) = -3 \]
So the other point is \( (-1, -1, -3) \).

### Final Points
The points on the line through A and B at a distance of 14 units from the midpoint M are:
1. \( (7, 11, 21) \)
2. \( (-1, -1, -3) \)