# Homework Help: Hopelessly confused with vectors

1. Sep 7, 2008

### Boogeyman

1. The problem statement, all variables and given/known data
In the diagram ABCD is a square, P and Q are the midpoints of DC and BC respectively. Show that PA + QA = 1.5 CA
check attachment for diagram

2. Relevant equations
none (i think?)

3. The attempt at a solution
PA+QA=3/2 CA
=PC+CA+QC+CA
=PA+AC+QC+2CA
=PA+AC+QA+AC+2AC
=PA+QA+4AC

and I'm stuck here. I also would like to know if there is a difference between prove
and show

Secondly is anyone could give some tips how to deal with these problems with proofs (also in trigonometry) it would be greatly appreciated. I am pretty much running through the dark with these problems.

P.S I don't know how to get the little arrows above the letters to show they are vectors, if anyone could tell me how...

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2. Sep 7, 2008

### danago

The way i would do it would be to let the length AD be represented by the vector a, and let DC be represented by b. Then try writing PA, QA and CA all in terms of only a and b, and see if you can see what to do next.

3. Sep 7, 2008

### edziura

I assume the notation means for example PA is a vector directed from point P toward point A. Start by writing vector sums for PA and QA in terms of CA and other vectors. So (again, these are vector equations)

PA = PC + CA and similarly QA = QC + CA

PA + QA = PC + QC + CA + CA = PC + QC + 2CA

If we are correct so far, we suspect that PC + QC must someow be equal to -0.5 CA.

Let O be the center of the square. Then consider the square OQCP. The diagonal from lower left to upper right of this square is the vector 0.5AC (note AC, not CA). Also from this square one can see that

PC + QC = 0.5AC

Now recall that, for example, AC = - CA, so that 0.5AC = - 0.5 CA, so

PC + QC = 0.5AC = -0.5CA

and we have what we need. There's probably a more direct proof, but this one works, and it makes use of vectors.

4. Sep 7, 2008

### Boogeyman

See this is as much as I suspected but I did not know for sure how PC + QC = 0.5AC. I know how to add a vector say QC + CR which would be equal to QR, but how do you add vectors like that i.e PC + QC?

Secondly, how can we assume the origin, O, is at the centre of the square?

5. Sep 7, 2008

### edziura

Secondly, how can we assume the origin, O, is at the centre of the square?

We don't need a coordinate system for this problem. Point O is just a way to label the center of the square.

See this is as much as I suspected but I did not know for sure how PC + QC = 0.5AC. I know how to add a vector say QC + CR which would be equal to QR, but how do you add vectors like that i.e PC + QC?

Slide PC to position OQ. Slide QC to position PO. The two vectors are now two sides of a parallelogram whose diagonal is their sum and = 0.5AC.

6. Sep 7, 2008

### Boogeyman

But wouldn't it be QC to OP? I know that the diagonal is the sum, but I still don't understand how the sum could be 0.5AC, or how we even obtain the value for the diagonal.

7. Sep 7, 2008

### edziura

But wouldn't it be QC to OP? I know that the diagonal is the sum, but I still don't understand how the sum could be 0.5AC, or how we even obtain the value for the diagonal.

Yes it would. The diagonal goes from point O, the midpoint of AC (since O is at the center of the square) to point C, and its length is 1/2 the length of the diagonal (AC) of the original square. Thus PC + QC = 0.5 AC

8. Sep 8, 2008

### Boogeyman

OH RIGHT..now I see. Wow vectors are really interesting once you understand.

Another question I have is about position vectors. Sometimes the question asks you about a position vector A eg. 2a - 3b relative to the origin, O, so then it's like OA = 2a - 3b. But then some ask you questions about the position vector alone

Eg. the position vector of A and B are a - 3b and 2a + 5b respectively. Find the position of the mid pt. of AB.

So if you were to draw a vector triangle how does it look? How do you work a question like this?

9. Sep 8, 2008

### edziura

See attached document.

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