Find the points on the line through A(-5,-4,1) and B(2,3,5) that is twice as for from A as from B.

Answer

To find the points on the line through A(-5, -4, 1) and B(2, 3, 5) that is twice as far from A as from B, we can follow these steps:

### Step 1: Understand the Problem
We need to find a point P on the line segment AB such that the distance from A to P is twice the distance from P to B. This can be expressed as:
\[ AP = 2 \cdot PB \]
This implies that the ratio of the distances is:
\[ \frac{AP}{PB} = 2 \]
This means that the point P divides the line segment AB in the ratio 2:1.

### Step 2: Identify the Coordinates of Points A and B
Let:
- A = (-5, -4, 1)
- B = (2, 3, 5)

### Step 3: Use the Section Formula
The section formula states that if a point P divides the line segment joining points A(x1, y1, z1) and B(x2, y2, z2) in the ratio m:n, then the coordinates of P are given by:
\[ P\left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}, \frac{mz_2 + nz_1}{m+n} \right) \]
In our case, m = 2 and n = 1.

### Step 4: Substitute the Values into the Section Formula
Using the coordinates of A and B:
- \( x_1 = -5, y_1 = -4, z_1 = 1 \)
- \( x_2 = 2, y_2 = 3, z_2 = 5 \)

Now substitute into the section formula:
\[ P_x = \frac{2 \cdot 2 + 1 \cdot (-5)}{2 + 1} = \frac{4 - 5}{3} = \frac{-1}{3} \]
\[ P_y = \frac{2 \cdot 3 + 1 \cdot (-4)}{2 + 1} = \frac{6 - 4}{3} = \frac{2}{3} \]
\[ P_z = \frac{2 \cdot 5 + 1 \cdot 1}{2 + 1} = \frac{10 + 1}{3} = \frac{11}{3} \]

### Step 5: Write the Final Coordinates of Point P
Thus, the coordinates of point P are:
\[ P\left(-\frac{1}{3}, \frac{2}{3}, \frac{11}{3}\right) \]

### Summary of the Solution
The point on the line through A and B that is twice as far from A as from B is:
\[ P\left(-\frac{1}{3}, \frac{2}{3}, \frac{11}{3}\right) \]