Hello, hymnseeker!

The position vectors of points $\displaystyle A$ and $\displaystyle B$ relative to an origin $\displaystyle O$

are: .$\displaystyle \vec A \:=\:\langle 5,4,1\rangle,\;\vec B \:=\:\langle -1,1,-2\rangle$

Find the position vector of point $\displaystyle P$ which lies on $\displaystyle AB$ produced so that: $\displaystyle AP\:=\:2\!\cdot\!BP$

My strategy was to calculate AB, then take 1/3 of this ... .no Did you make a sketch?

Code:

: ← - - - - 2 - - - - → :
*-----------*-----------*
A B ← - 1 - → P

As you can see, $\displaystyle B$ is the *midpoint* of $\displaystyle AP.$