Finding the unknown cordinates of a point on a vector

[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:PBSolution:

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.

 

[Dick]

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:PBSolution:

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]

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]

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]

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]

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]

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.