Question

If $\mathbf{a}$ and $\mathbf{b}$ are position vectors of point A and point B, respectively, find the position vector of point C on $\overrightarrow{BA}$ such that $BC = 3BA$.

Hint:

When a point divides a vector in a given ratio, use the concept of weighted averages to find the position vector.

Answer 1

Let $\mathbf{a}$ and $\mathbf{b}$ be the position vectors of points A and B, respectively. The vector $\overrightarrow{BA}$ is calculated as $\mathbf{a} - \mathbf{b}$. We need to find the position vector $\mathbf{c}$ of point C such that the magnitude of $\overrightarrow{BC}$ is three times the magnitude of $\overrightarrow{BA}$, i.e., $BC = 3BA$. The vector $\overrightarrow{BC}$ is given by $\mathbf{c} - \mathbf{b}$. Therefore, we have the equation: \[ \mathbf{c} - \mathbf{b} = 3(\mathbf{a} - \mathbf{b}). \] Expanding and rearranging this equation yields: \[ \mathbf{c} - \mathbf{b} = 3\mathbf{a} - 3\mathbf{b}, \] which simplifies to: \[ \mathbf{c} = 3\mathbf{a} - 2\mathbf{b}. \] This is the position vector of point C.