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Finding the unknown cordinates of a point on a vector

by EmilyHopkins
Tags: cordinates, point, unknown, vector
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EmilyHopkins
#1
Dec18-12, 09:55 AM
P: 8
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.
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Dick
#2
Dec18-12, 10:43 AM
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Quote Quote by EmilyHopkins View Post
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 = (1-t)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.
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.
EmilyHopkins
#3
Dec18-12, 11:02 AM
P: 8
Quote Quote by Dick View Post
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.
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

Dick
#4
Dec18-12, 11:07 AM
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Finding the unknown cordinates of a point on a vector

Quote Quote by EmilyHopkins View Post
Well if AP and PB are parallel then their cross product is 0.
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.
EmilyHopkins
#5
Dec18-12, 11:20 AM
P: 8
Quote Quote by Dick View Post
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.

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.
Dick
#6
Dec18-12, 11:31 AM
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Quote Quote by EmilyHopkins View Post
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 + j)
0 = - (t-3)(t-7)k + (t-1) k
0 = (3-t)(t-7) k + (t-1)k
0= (3t -t^2-21+7t)k + (t-1)k
0= (10t -t^2 -21)k + (t-1)k
0= (11t - t^2 -22)k

11t - t^2 -22 =0

a= -1, b= 11, c=--22

t = -11 ▒ (121 -4(-1)(-22))1/2/2(-1)

t= -1.72 or t= 12.7 ???
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.
Dick
#7
Dec18-12, 11:33 AM
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Quote Quote by EmilyHopkins View Post
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.
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.


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