Problem with vectors

[Icheb]

I have a problem with the following:

 

a and b are non-parallel vectors. Choose two numbers x and y so that

 

3r_1 = 2r_2

 

can be applied to the following vectors

 

r_1 := (x + 4y) a + (2x + y + 1) b

r_2 := (y - 2x + 2) a + (2x - 3y - 1)b

 

Now, I have no idea what to do with the vectors. I could solve it if it were just for x and y without a, b and r, but I don't know whether to just drop the vectors, use some vectors i choose myself or if I have to calculate it with a and b left there as a variable.

 

The solution is x = 2 and y = -1 btw, but even with that information I can't deduce the proper way of calculating it.

 

Can someone give me a hint as to how to approach this?

[timo]

I haven´t tried solving for the solution but atm I don´t see much problem with the question - except perhaps for one tricky thing you´ll possibly encounter many times again.

I´d try solving the question the following way:

You are given [math] 3 \vec r_1 - 2 \vec r_2 = \vec 0 [/math]

Plug in the definitions of r1 and r2 and rewrite it to the following form:

[math] A(x,y) \vec a + B(x,y) \vec b = \vec 0 [/math] where A(x,y) and B(x,y) are some terms dependent on x and y (the prefactors).

Now, here comes the trick: Since a and b are not colinear, the only option way to have a linear combination of them resultng in the zero-vector is having both coefficients equal zero: [math] 0 \vec a + 0 \vec b = \vec 0 [/math].

Therefore, both A(x,y) and B(x,y) have to equal zero:

A(x,y)=0

B(x,y)=0

 

That´s two equations for two unkowns x and y. Solve them.

 

Like I said, I didn´t actually solve that problem so perhaps there´s some trapdoors hidden in it - but I doubt it.

[Icheb]

Would it be possible that you explain why A(x,y) has to be 0?

The only solution I could think of was 3r_1 = 0 and 2r_2 = 0 and then calculate x and y, which is of course wrong. But I just don't understand why A(x,y) and B(x,y) have to be 0 in order for it to be correct.

[timo]

Would it be possible that you explain why A(x,y) has to be 0?

I could, but since I already explained that in my first post I don´t think that this actually is your problem. Again: The only possibility for a linear combination of two non-parallel vectors a and b to equal the zero-vector is the linear combination 0*a + 0*b.

 

The only solution I could think of was 3r_1 = 0 and 2r_2 = 0 and then calculate x and y, which is of course wrong.

This is what I think your problem in understanding is, actually. You are not given the equations 3 r1 = 0 and 2 r2 = 0 but the equation 3 r1 = 2 r2. This is equivalent to the equation 3 r1 - 2 r2 = 0 which was my starting point in above.

 

But I just don't understand why A(x,y) and B(x,y) have to be 0 in order for it to be correct.

Because they are the coefficients of the linear combination mentioned in above - "A(x,y)" and "B(x,y)" are just names I gave to them. For example, A(x,y) = 3(x + 4y) - 2(y - 2x -2).

[Icheb]

I could, but since I already explained that in my first post I don´t think that this actually is your problem.

Well I wouldn't be asking you for this if I understood the first explanation. Thanks anyways.

[timo]

I can repeat it for a 3rd time if you want. I´m just not sure that this specific question is your problem. I´d rather expect that you´re confused by that x and y are not entries of a vector but only some numbers without much "meaning".

So for a 3rd time (and again with different terms): If two vectors are non-parallel, then they are lineary independent. The DEFINITION of linear independence of vectors is that the only way a linear combination of them becomes zero is the trivial one: Each coefficient of the linear combination is zero.

It´s not that I don´t want to help you - it´s just that I don´t really know where your problem lies. Perhaps if you posted step-by-step how far you got before running into a problem that would help.

The first step is using 3 r1 = 2 r2 and rewriting it to the form A*a + B*b = 0.