1. 3D vector question

The position vectors of the points A and B relative to an origin O are 5i + 4j + k, -i + j - 2k respectively. Find the position vector of the point P which lies on AB produced such that AP=2BP.
My strategy was to calculate AB, then take 1/3 of this and add it to the position vector for B. However this returns the incorrect answer.

Any ideas?

2. Originally Posted by hymnseeker
My strategy was to calculate AB, then take 1/3 of this and add it to the position vector for B.
The equation $\displaystyle \overrightarrow {AP} = 2\overrightarrow {BP} \, \Rightarrow \,A - B - P$ that is B is between A&P.
Your approach puts P between A&B.
Try this $\displaystyle \overrightarrow P = 2\overrightarrow {AB} + \overrightarrow {A}$. Does that work?

3. 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 $\displaystyle AP.$

4. I see...

Thanks for your help Plato and Soroban, I understand now.