position vector given points, but ratio between the two points?

• Feb 6th 2009, 10:43 PM
Random-Hero-
position vector given points, but ratio between the two points?
Use vectors to find the position vector of point 'P' if 'P' divides AB in the ratio 3:2 given A(-1,6,4) and B(4,1,-1).

This is what I'm thinking so far..
Point P is 3/5 the way along from A to B. When you divide something in the ratio 3:2, one part will be 3/5 of the total and the other will be 2/5 of the total.

coordinates of P are (x, y, z), the position vector of point P will be the vector (x - (-1), y - 6, z - 4). This vector has the same direction as AP and the same magnitude.

Am I on the right track here fellas?
• Feb 7th 2009, 12:51 AM
Bruce
First find the column vector AB = (5, -5, -5)

vector AP = 3/5 vector AB
vector AP = 3/5(5, -5, -5)
vector AP = (3, -3, -3)

So move 3, -3 and -3 from A (-1, 6, 4) to find the co-ordinates of P.
• Feb 7th 2009, 10:31 PM
Position vectors
Hello everyone -

You'll find that the answer is simply

$\tfrac{2}{5}\vec{a} + \tfrac{3}{5}\vec{b}$

where, of course $\vec{a}$ and $\vec{b}$ are the position vectors of A and B. Which simply says that P's position is two-fifths of A's plus three-fifths of B's.

Can you see why this is? As you were saying, you move to P by moving to A and then three-fifths of the way from A to B. So:

$\vec{p} = \vec{a} + \tfrac{3}{5}\vec{AB}$

$= \vec{a} + \tfrac{3}{5}(\vec{b} - \vec{a})$

$= \tfrac{2}{5}\vec{a} + \tfrac{3}{5}\vec{b}$

And you can generalise this, and say that if P divides AB in the ratio $\lambda:\mu$, then

$\vec{p} = \frac{\mu}{\lambda+\mu}\vec{a} + \frac{\lambda}{\lambda+\mu}\vec{b}$

(Notice how the $\lambda$ and $\mu$ have 'swapped' over.)

I hope you find this useful.