# Homework Help: Vector Problem finding a certain endpoint.

1. May 18, 2012

### charmedbeauty

1. The problem statement, all variables and given/known data

Let A=(2,3,-1) and B=(4,-5,7). Find the midpoint of A and B. Find the point Q on the line through A and B such that B lies between A and Q and BQ is three times as long as AB

2. Relevant equations

3. The attempt at a solution

So to find the midpoint of A and B.

MA+B=(A+B)/2

RHS= (6,-2,6)/2

MA+B=(3,-1,3)

Next I said that the the midpoint of Q and A is 3 × (MA+B)

Hence MQ+A= (9,-3,9)

So now we have the midpoint and one endpoint of the line QA.

so Q+A=2(MQ+A)

so RHS= (18,-6,18)

Hence Q+2=18
Q+3=-6
Q-1=18

so Q=(16,-9,19)

But this was wrong I don't see why?

2. May 18, 2012

### HallsofIvy

Yes, that is correct.

Why did you change from calling it "MA+B" to "RHS"?

Why would you think that? It certainly is NOT true. To take a simple case, suppose A= 0 and B= 1 on a number line. The distance from A to B is 1 and so the distance from B to Q must be 3. That is, Q= 1+ 3= 4. The midpoint of AB is 1/2 and the midpoint of AQ is 2. 2 is NOT 3 times 1/2.

Because you entire logic is incorrect. The midpoint of AQ is NOT three times the midpoint of AB.

Instead, the length of BQ is three times the length of AB which means the length of AQ= AB+BQ is four times the length of AB. Using "similar triangles" that means that Qx= Ax+ 4(B[subx[/sub]- Ax)= 2+ 4(4- 2)= 2+ 4(2)= 10, etc.

3. May 18, 2012

### charmedbeauty

Ohh yes it is for times the magnitude... so can I use the same method and say the midpoint of the new line is 4 times the original midpoint?