Given the position vectors of points A (1, -1), B (5, -3), and C (11, -6), show that the vectors AB and BC are parallel.

The position vectors are:

To find the vector $\vec{AB}$, subtract the position vector of A from the position vector of B:

To find the vector $\vec{BC}$, subtract the position vector of B from the position vector of C:

Two vectors are parallel if one is a scalar multiple of the other. Let's check if $\vec{BC} = \lambda \vec{AB}$ for some scalar $\lambda$.

From the x-components: $6 = 4\lambda \implies \lambda = \frac{6}{4} = \frac{3}{2}$
From the y-components: $-3 = -2\lambda \implies \lambda = \frac{-3}{-2} = \frac{3}{2}$
Since $\lambda = \frac{3}{2}$ is consistent for both components, we have $\vec{BC} = \frac{3}{2} \vec{AB}$.

Since $\vec{BC}$ is a scalar multiple of $\vec{AB}$, the vectors are parallel. Both vectors share a common point B. Therefore, points A, B, and C lie on the same straight line, meaning they are collinear.