# Homework Help: Finding the unknown cordinates of a point on a vector

1. Dec 18, 2012

### EmilyHopkins

Relative to the origin O, the position vectors of two points A and B are (1,4) and (7,1) respectively. Give that the point P (t,t+1) is on AB find

1) AP and BP in terms of t

2) Find the value of t and hence the ratio AP:PB

Solution:

1) AP= (-i - 4 j) + ti + (t+1)J = (t-1)i + (t-3)j
BP= -7i -j + ti + (t+1)j =(t-7)i + (t)j

2) I have no idea how to find t.

Last edited: Dec 18, 2012
2. Dec 18, 2012

### Dick

AP and PB are parallel vectors, since P is on AB, right? How do you express two vectors being parallel in algebra? BTW your expression for AP has a sign mistake.

Last edited: Dec 18, 2012
3. Dec 18, 2012

### EmilyHopkins

Well if AP and PB are parallel then their cross product is 0.

0 = AP X PB

0= ((t -1)i + (t-3)j) X (( T-7)i + tj)
0 = - (t-3)(t-7)k + (t-1)t k
0 = (3-t)(t-7) k + (t-1)tk
0= (3t -t^2-21+7t)k + (t-1)tk
0= (10t -t^2 -21)k + (t-1)tk
0= (10t -t^2 -21)k +(t^2 - t)k
0= (9t -21)k

9t-21=0
9t=21
t= 7/3

Last edited: Dec 18, 2012
4. Dec 18, 2012

### Dick

That would be one way to go if you extend them to three dimensional vectors. It's also true that if they are parallel then they are multiples of each other. k*AP=PB for some constant k.

5. Dec 18, 2012

### EmilyHopkins

Wouldn't that just introduce another unknown variable which would require us to have 2 equations in order to solve. I thought this route already but didn't bother going this way since I don't know the ratio, and hence the value of k.

Last edited: Dec 18, 2012
6. Dec 18, 2012

### Dick

You are being pretty sloppy here. (t-7)i + (t)j turned into (( T-7)i + j). Something missing. If you do this right the t^2 will cancel.

7. Dec 18, 2012

### Dick

Once you split into components it WILL turn into two equations in the two unknowns t and k. It's about the same amount of work to do it this way as to solve the cross product equation.