Help me with cross product question!


by stunner5000pt
Tags: cross, product
stunner5000pt
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#1
Oct29-04, 06:52 PM
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Find two vectors which are perpendicular, and one of them is twice the magnitude of the other. Also their sum should be (6,8).

First of all let the coords be (x,y) and (a,b)

then the magnitude of one of them is twice the other i .e . [tex] 2 \sqrt{x^2+y^2} = \sqrt{a^2+b^2} [/tex]
also their dot product is zero

[tex] (x,y) \cdot (a,b) = 0 -> xa + yb = 0 [/tex]

also the cross product is just the magnitudes of the vectors multiplied

[tex] (x,y) \times (a,b) = 2 (x^2 + y^2) [/tex]

But somehow i cannot get the fact that they add up to 6,8 to work into this framework i have setup

Is it simpler than i think it is??

Please help!!!!! Thank you in advance!
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Hurkyl
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Oct29-04, 07:01 PM
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Well, first off, you haven't written equations describing all of the conditions of the problem...
stunner5000pt
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#3
Oct29-04, 07:12 PM
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ok then one more equation i missed is [tex] (x,y) + (a,b) = (6,8) [/tex]

Fredrik
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Oct29-04, 07:28 PM
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Help me with cross product question!


The vectors are perpendicular: xa+yb=0
The magnitude of one of the vectors is twice the magnitude of the other: 4(x+y)=a+b
The sum is (6,8): x+a=6; y+b=8

I don't know what you're doing with that cross product. The cross product of two vectors is perpendicular to both vectors.
stunner5000pt
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#5
Oct29-04, 08:28 PM
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Quote Quote by Fredrik
The vectors are perpendicular: xa+yb=0
The magnitude of one of the vectors is twice the magnitude of the other: 4(x+y)=a+b
The sum is (6,8): x+a=6; y+b=8

I don't know what you're doing with that cross product. The cross product of two vectors is perpendicular to both vectors.
ok what am i supposed to do after writing the equations down (sorry i dont mean to be sarcastic, but i'm not getting a reasonable reply here!)
vsage
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Oct29-04, 09:00 PM
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What you ultimately want is from the four unknowns, a relationship of three to the other one. For example, a = 3x, b = 4x, y = 5x. It's not quite that simple in this case but that is about the best scenario you could get. I'll try and type out what I found. OK here is goes:

1. The sum of the vectors must be (6,8)
(a,b)+(x,y) = (6,8)
a+x = 6
b+y = 8
a+x+b+y=14


2. The two vectors must be perpendicular
[tex](a,b) \cdot (x, y) = 0[/tex]
[tex] ax + by = 0[/tex]

3. The magnitude of one is twice the magnitude of the other
[tex]\sqrt{x^2+y^2} = 2 \sqrt{a^2+b^2}[/tex]
vsage
#7
Oct29-04, 09:29 PM
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Hrm I wasn't that keen on what I said above so here's a little shove in the direction I would go to attack the problem. Use equations 1, 2 and 3 and solve for x:

1. x = 14-a-b-y
2. [tex]x=\frac {-by}{a}[/tex]
3. [tex]x = \sqrt{4(a^2+b^2)-y^2}[/tex]

Set equation 1 = equation 2 and solve for another variable. Set equation 2 equal to equation 3 and solve for that same variable. You will now have two equations with two unknowns relating to another variable. Set those two equal and solve for another variable. Simple substitution back into your many derived equations will give you a relationship of three variables to the other one.
Fredrik
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#8
Oct29-04, 10:30 PM
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Quote Quote by stunner5000pt
ok what am i supposed to do after writing the equations down (sorry i dont mean to be sarcastic, but i'm not getting a reasonable reply here!)
You did get a reasonable reply. You seemed to be having problems finding the right equations (and you were saying some strange things about cross products), so I tried to help you with that. I thought that was what you needed to get started. I will not solve this problem completely for you, because that's not what we do here, but I will tell you a few more things.

There are four unknowns, but there are also four equations, so it should be possible to solve the system completely for a, b, x and y. (If there had been only three equations we would have had to settle for a way to express three of the unknowns as functions of the fourth). It can be difficult to figure out how to solve a non-linear system of equations such as this one, but if you play around with the equations for a while, you may find a trick that simplifies things. One such trick is use all of the equations to express (x+a)+(x+b) in two different ways. The result is an equation that will tell you the norms of the vectors.

When you have the norms, you can proceed e.g. like this: 0=xa+yb=a(6-a)+b(8-b)=... If you use the result about what a+b is, this equation simplifes to a relationship between a and b that tells you the "direction" of the vector (a,b). A similar calculation that starts with the same equation will tell you the "direction" of (x,y).


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