# Calc 3, vector problem

1. Sep 19, 2011

### Allenman

I mostly just want to know if I did this correctly... And if not, where I went wrong.

1. The problem statement, all variables and given/known data
Of $\overline{U}$=(-2, 6) is the vector $\overline{AB}$ and the midpoint of the line segment from A to B has the coordinates (2,1). Find A and B.
Let
$\overline{A}$=(a1, a2)
$\overline{B}$=(b1, b2)

2. Relevant equations

3. The attempt at a solution
$\overline{AB}$=(a1, b1) + t(b1-a1, b2-a2)

rearranging I got:
$\overline{AB}$=(1-t)(a1, b1) + t(a2, b2)

So when t=0 I have the coordinates (a1 and b1)
and when t=1 I have the coordinates (a2 and b2)

So plugging the numbers:
$\overline{AB}$=(-2,6) + t(-2 - 2, 6 - 1)
=(1-t)(-2, 6) + t(2, 1)
$\overline{A}$=(-2, 2)
$\overline{B}$=(6, 1)

Did I do it right? or am I way off?... lol

2. Sep 19, 2011

### issacnewton

allen,

i think something is wrong. if t=0 , you should get point A, the coordinate
$(a_1,b_1)$ is not point A.

I will do this as following. let O be the origin. and let the coordinates of A and B be
(a1,a2) , (b1,b2). let the M be midpoint of
the segment AB , then

$$\vec{AB}=\vec{OB}-\vec{OA}=(b_1,b_2)-(a_1,a_2)=(b_1-a_1,b_2-a_2)$$

$$\vec{AM}=\vec{OM}-\vec{OA}=(2,1)-(a_1,a_2)=(2-a_1,1-a_2)$$

$$\vec{MB}=\vec{OB}-\vec{OM}=(b_1,b_2)-(2,1)=(b_1-2,b_2-1)$$

now realize that vectors AM and MB are the same since they have same magnitude and direction.

$$\vec{AM}=\vec{MB}$$

so you get 4 equations and 4 unknowns . solve it. I got

$$A=(3,-2) \; , B=(1,4)$$

just to confirm the answer, use midpoint formula for A and B and you get back the coordinate of point M.

3. Sep 19, 2011

### Allenman

Thank you very much! You're awesome! :)