Vector ratio problem

Apr 2010
60
0
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
 
Dec 2009
872
381
1111
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,

Hope this will help you. If you need any clarifications please don't hesitate to reply me.
 

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Apr 2010
60
0
I'm still a bit confused with the ratio..can you solve the problem and show me?
 
Dec 2009
872
381
1111
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.

Hope this will help you.
 

Grandad

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Dividing a line segment externally

Hello 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.

Grandad
 

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