1. ## Vector ratio problem

The point C has position vector (2 3) and point D has position vector (1 2).Find the position vector of the point which divides CD in the ratio 4:-3

I recognized that the minus sign in 4:-3 indicates that it divides the line externally.However,i'm not able to get the ratios right when solving the problem.Can anyone show me with a diagram,the way the ratio is distributed in the line? (< i guess this is where i get it wrong?the ratio part)
Thanks

2. Originally Posted by kandyfloss
The point C has position vector (2 3) and point D has position vector (1 2).Find the position vector of the point which divides CD in the ratio 4:-3

I recognized that the minus sign in 4:-3 indicates that it divides the line externally.However,i'm not able to get the ratios right when solving the problem.Can anyone show me with a diagram,the way the ratio is distributed in the line? (< i guess this is where i get it wrong?the ratio part)
Thanks
Dear kandyfloss,

3. I'm still a bit confused with the ratio..can you solve the problem and show me?

4. Originally Posted by kandyfloss
I'm still a bit confused with the ratio..can you solve the problem and show me?
Dear kandyfloss,

Can you explain about what confuses you? If the point that divides CD externally is E(x,y), you can write,

$\displaystyle 1=\frac{x+(3\times{2})}{4}$ ; considering D divides CE internally

$\displaystyle 2=\frac{y+(3\times{3})}{4}$

Hence you can find the point E.

5. ## Dividing a line segment externally

Hello kandyfloss
Originally Posted by kandyfloss
The point C has position vector (2 3) and point D has position vector (1 2).Find the position vector of the point which divides CD in the ratio 4:-3

I recognized that the minus sign in 4:-3 indicates that it divides the line externally.However,i'm not able to get the ratios right when solving the problem.Can anyone show me with a diagram,the way the ratio is distributed in the line? (< i guess this is where i get it wrong?the ratio part)
Thanks
You're right in saying that the point divides the line segment $\displaystyle CD$ externally. If $\displaystyle CP:PD = 4 :-3$, then the ratio of the distances $\displaystyle CP$ and $\displaystyle PD$ is $\displaystyle 4:3$, but they are in opposite directions (hence the minus sign). So, from $\displaystyle C$ we shall go out $\displaystyle 4$ 'lengths' to get to $\displaystyle P$ and then come back $\displaystyle 3$ 'lengths' to $\displaystyle D$.

Take a look at the diagram I've attached. You'll see that:
$\displaystyle CP:PD = 4:-3$
Now you probably know that the position vector of the point dividing the line joining the points with position vectors $\displaystyle \vec a$ and $\displaystyle \vec b$ in the ratio $\displaystyle \lambda:\mu$ is:
$\displaystyle \frac{\mu\vec a + \lambda \vec b}{\lambda + \mu}$
So the point that divides $\displaystyle CD$ in the ratio $\displaystyle 4:-3$ has position vector:
$\displaystyle \frac{-3\vec c + 4\vec d}{4-3}$
$\displaystyle =-3\binom23+4\binom12$

$\displaystyle =\binom{-6}{-9}+\binom48$

$\displaystyle =\binom{-2}{-1}$

I hope that clears things up.