Method for proofs involving vectors and dot products?

1. Jan 5, 2014

Ascendant78

Ok, I'm going to be taking calc III next week, so I wanted to get a head-start by doing the MIT multivariable calculus opencourseware. While most of the material was easy, these proofs are really killing me. Here are two examples:

Ex.1: Using vectors and dot product show the diagonals of a parallelogram have equal lengths if and only if it’s a rectangle.

Ex. 2: The median of a triangle is a vector from a vertex to the midpoint of the opposite side. Show the sum of the medians of the triangle.

Now, what I am wondering is if there's a specific type of method you utilize to prove these types of things? I feel like every time I try one on my own, I end up hitting a brick wall. I will figure out a handful of different restrictions and such, but I never seem to be able to tie up the loose ends.

I don't need the answers to either example as they have them online. I am just lost as to a method to figure these types of problems out. If anyone has any strategies, I'm all ears.

2. Jan 6, 2014

Staff: Mentor

There's no specific method. Often it boils down to labeling your diagram and constructing some vector additions based on the problem that you have and then reduce them algebraically to get the result. Sometimes it helps to work the problem from both ends and hope they meet.

and for other related problems:

I ran into a similar problem when learning vectors again where I doubted the sin of the sum of two angles and had to find the proof because I couldn't do it. It involved some clever labels and additional lines with the final result showing that the addition of two line segments were the same as sin(a+b) line segment and then I went duh...

3. Jan 6, 2014

tiny-tim

Hi Ascendant78!
Just translate the question into vector notation.

eg, if the sides of the parallelogram are a and b, what are the lengths of the diagonals?

subtract them, and translate back into english!

Similarly, if two sides of the triangle are a and b, what are the three medians?

Show us what you get

4. Jan 6, 2014

Ascendant78

The proofs that were given to use from MIT were far more convoluted. I also feel like they only skimmed over what we needed to know, then had us applying the concepts in a way they hadn't explained yet. Can't really blame MIT though as they do caution users that you can run into that with the OCW sometimes. Just makes the learning take 10x longer without the explanations first before the problems. Anyway, thanks again, I really appreciate the help.

5. Jan 6, 2014

Ascendant78

Thanks for the feedback. I know now that a part of my problem is not having a solid enough understanding of vectors to manipulate them with comfort yet. I think I just need to spend more time on these proofs to really get a solid understanding. I also think that I need to use my courses textbook along with the MIT OCW materials, as they don't quite seem complete. When I went back last night and read from my textbook and compared that to what I went over in the OCW, it all made a lot more sense.

6. Jan 7, 2014

Ascendant78

Ok, well I've been working on these for a bit now and still having difficulty. I have to say I'm really frustrated because the professors at our college told us Calc III is what most students consider the easiest, but I'm finding these proofs really difficult. My problem is I don't necessarily know what accounts as proof in most cases - I can't tell exactly when what I've provided suffices.

Do these vector and dot product proofs typically take a while for a person to wrap their mind around? I've never had this much trouble grasping a mathematical concept and am really frustrated.

7. Jan 7, 2014

Ascendant78

Ok, here is an example of what I'm talking about (*EDIT* - problem asked you to prove using vectors):

In this proof, I started with defining the midpoint of each diagonal as two distinct points. Then, I sought to prove that both midpoints are located at the same point in space, which would mean that they bisect each other. Then, I used the "B-A = C-D" and "C-B = D-A" based on the definition of a parallelogram. After that, I simply manipulated variables in each of the equations and used substitution to cancel until both P and Q were equal to the same values.

While this makes sense to me and I feel I applied both vectors and the definition of a parallelogram to prove it, I am not sure if this is adequate or not? If it is not, can you please tell me why?

Also, is this something we really need to get down solid for a physics major? I have already spent hours and hours on it and don't want to keep putting time into it if it is not something I'll see much of later on.

Last edited: Jan 7, 2014
8. Jan 7, 2014

Staff: Mentor

You don't need to define two separate points only the intersection of the diagonals point and then prove that the line segments are equal in length.

9. Jan 8, 2014

tiny-tim

Hi Ascendant78!
Yes, your proof is adequate, but more complicated than it needed to be.

You probably feel it's not very intuitive.

For proofs like this, it usually helps to find a notation that makes everything shorter and clearer.

In this case, you could start by saying let the origin be at one corner of the parallelogram, and let the two adjacent sides be a and b.

Then the two diagonals are … and … ?

And the midpoints are at … ?​

This a and b approach can be applied to any parallelogram or triangle problem, and has the advantage that it shows clearly how to apply the usual "parallelogram rule" for adding and subtracting vectors.

10. Jan 8, 2014

Ascendant78

Ok, I see what you mean. Thanks again. Also, you are right that I don't feel these are intuitive at all. I feel like I just start throwing out various vectors based on the figure and the restrictions that go with it, then I start tossing out vectors in attempts to unite them with the restrictions. I don't like that I don't feel like there is much of a method behind it beyond me just going from point a and point b and hoping I can get them to meet somewhere in between eventually.

11. Jan 8, 2014

tiny-tim

have another go now at your original two examples …

maybe something will click!