Solution:

\(2+1=3\), so we are dividing the line into thirds.

Earlier we calculated the vector \(\vec{AB}\) to be \(4 \vec{\imath}-5 \vec{\jmath}\)

Because of the way in which vectors can be added, \( \vec{OC} = \vec{OA} + \vec{AC} = \vec{OA} + \frac 13 \vec{AB} \)

(In other words, going from \(O\) to \(C\) is the same as going from \(O\) to \(A\) and then going from \(A\) to \(C\).)

This means that \[\vec{OC}=\begin{bmatrix}1\\4 \\\end{bmatrix}+ \frac 13 \begin{bmatrix}-4\\-5 \\\end{bmatrix} = \begin{bmatrix}1- \frac 43\\4- \frac 53\\\end{bmatrix}\]

\[= \begin{bmatrix} -\frac 13 \\ \frac 73 \\\end{bmatrix}\]

You can use the diagram to check that this answer makes sense - the green arrow is just a little bit to the left of the x-axis, corresponding to the negative x-coordinate and more than halfway up to \(A\) corresponding to the positive y-coordinate.

Problems using Vectors