Vector Problem finding a certain endpoint.

[charmedbeauty]

Homework Statement

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

Homework Equations

The Attempt at a Solution

So to find the midpoint of A and B.

M_A+B=(A+B)/2

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

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

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

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

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

so Q+A=2(M_Q+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?

 

[HallsofIvy]

Homework Statement

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

Homework Equations

The Attempt at a Solution

So to find the midpoint of A and B.

M_A+B=(A+B)/2

Yes, that is correct.

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

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

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

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

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.

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

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

so Q+A=2(M_Q+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?

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 Q_x= A_x+ 4(B[subx[/sub]- A_x)= 2+ 4(4- 2)= 2+ 4(2)= 10, etc.

 

[charmedbeauty]

Yes, that is correct.


Why did you change from calling it "M_A+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 Q_x= A_x+ 4(B[subx[/sub]- A_x)= 2+ 4(4- 2)= 2+ 4(2)= 10, etc.

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?