Dot Product Proof: Prove |a|^2 = 2|b|^2

[EmilyHopkins]

Homework Statement

The point P and Q have postion vectors a + b, and 3a - 2b respectively, relative to the origin O.Given that OPQR is a parallelogram express the vector PQ and PR in terms of a and b. By evaluating two scalar products show that if OPQR is a square then |a |² = 2 |b |²

The Attempt at a Solution

OP = a + b

OQ = 3a -2b

PQ = OR= PO + OQ = -OP +OQ = -(a + b) + (3a - 2b) = 2a - 3b

PR = PO + OR = -OP + OR = -(a + b) + (2a - 3b) = a - 4b

So now the question says use two scalar products to show |a |² = 2 |b |². I'm assuming since the question ask for these two vectors in terms of a, and b that you will have to utilize it to get the result. So I drew out the square for a visualization.

So since its a square it means the dot product of

OP.PQ= 0
(a + b)(2a - 3b) = 2a² -ab -3b² = 0

ab = 2a² -3b²

PR.OQ= 0 (Since their perpendicular)
(a-4b)(3a - 2b) =0
3a² - 14ab +8b²=0
3a² -14(2a² -3b²) +8b²=0
3a² -28b² +42a² + 8b²=0
-25a²+50b²=0
25a²=50b²
a²= 2b²

Is their a faster way to work it or is this correct ?
?

 

[Simon Bridge]

What is the relationship between a and b? Are they just two arbitrary, non parallel, vectors? Orthogonal unit vectors? What?

Does the diagram come with the question or is it one you have drawn assuming that the parallelogram is a square? Note: the angles in a parallelogram do not have to be 90 degrees. Q does not have to be on the opposite corner to O. (Off the question alone, I'd have taken OP and OQ to be adjacent sides, and OR=OP+OQ.)

 

[EmilyHopkins]

What is the relationship between a and b? Are they just two arbitrary, non parallel, vectors? Orthogonal unit vectors? What?

Does the diagram come with the question or is it one you have drawn assuming that the parallelogram is a square? Note: the angles in a parallelogram do not have to be 90 degrees. Q does not have to be on the opposite corner to O. (Off the question alone, I'd have taken OP and OQ to be adjacent sides, and OR=OP+OQ.)

I'm assuming a and b are two non-parallel arbitrary vectors, as the question never specified what their relationship was. The diagram was used to visualize and help solve the second part of the question where they stated that if the parallelogram OPQR was a square, show by using two scalar products that |a |2 = 2 |b |2 .

 

[Simon Bridge]

Oh I missed the "if it were square" part.
The relationship means that a is the hypotenuse of a 1-1-root-2 triangle.

If you swap the positions of Q and R on your square, will it still fit the description?

In the following:

(a + b)(2a - 3b) = 2a² -ab -3b² = 0

you only expanded to three terms;
since a and b are vectors, a.a = a² is a little ambiguous;
I think you need to choose a notation that distinguishes between the length of a vector and the vector itself. In the above case:
(a+b).(2a-3b)=2a.a+2b.a-3a.b-3b.b=2|a|²-3|b|²+... how would you handle the mixed dot products? Is a.b the same as b.a?

Other than that - I think you have the actual method intended.
The only wrinkle remaining is the thing about the position of Q and R.
Does it make a difference?