Icheb Posted January 1, 2006 Share Posted January 1, 2006 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? Link to comment Share on other sites More sharing options...
timo Posted January 1, 2006 Share Posted January 1, 2006 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. Link to comment Share on other sites More sharing options...
Icheb Posted January 1, 2006 Author Share Posted January 1, 2006 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. Link to comment Share on other sites More sharing options...
timo Posted January 2, 2006 Share Posted January 2, 2006 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). Link to comment Share on other sites More sharing options...
Icheb Posted January 2, 2006 Author Share Posted January 2, 2006 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. Link to comment Share on other sites More sharing options...
timo Posted January 2, 2006 Share Posted January 2, 2006 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. Link to comment Share on other sites More sharing options...
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