The points $A(1, -2, 3)$, $B(3, 0, 1)$ and $C(4, 1, 0)$ are collinear. How does $C$ divide $AB$?

Explanation

1. 01

   $\overrightarrow{AC} = 3\hat{i} + 3\hat{j} - 3\hat{k}$ and $\overrightarrow{CB} = -\hat{i} - \hat{j} + \hat{k}$, so $\overrightarrow{AC} = -3\overrightarrow{CB}$.

2. 02

   The lengths are in the ratio $AC : CB = 3 : 1$, and the minus sign puts $C$ outside $AB$, beyond $B$: external division.

3. 03

   Check: $\frac{3(3, 0, 1) - (1, -2, 3)}{3 - 1} = (4, 1, 0)$.