# 3D vector question

• Apr 11th 2009, 01:43 PM
hymnseeker
3D vector question
Quote:

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?
• Apr 11th 2009, 03:32 PM
Plato
Quote:

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?
• Apr 11th 2009, 04:14 PM
Soroban
Hello, hymnseeker!

Quote:

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 $\displaystyle AP.$

• Apr 12th 2009, 12:05 AM
hymnseeker
I see...

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