- #1

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λx + (a cross x) = b

I have worked it down as far as

x.(λb + (b cross a)) = |b|^2

by taking the dot product with b of both sides.

But is there any way I can now solve this equation for x?

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- Thread starter elevenb
- Start date

- #1

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λx + (a cross x) = b

I have worked it down as far as

x.(λb + (b cross a)) = |b|^2

by taking the dot product with b of both sides.

But is there any way I can now solve this equation for x?

- #2

Simon Bridge

Science Advisor

Homework Helper

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Starting from: $$\lambda \vec x + \vec a \times \vec x = \vec b$$

You managed to get: $$\vec x \cdot \left( \lambda\vec b + \vec b \times \vec a \right)=|\vec b|^2$$

And you want to know how to solve that for ##\vec x## ??

This is a relation of form ##\vec x \cdot \vec y = k##

... how would you normally solve that for ##\vec x##?

- #3

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I don't know how to solve it for x? Is it possible?

- #4

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No. You have but one scalar expression here. The vector x has three components. You need to find two more equations before you can solve for x.

λx + (a cross x) = b

I have worked it down as far as

x.(λb + (b cross a)) = |b|^2

by taking the dot product with b of both sides.

But is there any way I can now solve this equation for x?

- #5

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I thought that? But I don't know how to work this equation around so I can solve it that way?

- #6

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Hint: You said "I have worked it down as far as x.(λb + (b cross a)) = |b|^2 by taking the dot product with b of both sides." What made you choose b? There's one other obvious choice, and a not so obvious third choice.

Hint #2: x is not one of those choices.

- #7

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Hint: You said "I have worked it down as far as x.(λb + (b cross a)) = |b|^2 by taking the dot product with b of both sides." What made you choose b? There's one other obvious choice, and a not so obvious third choice.

Hint #2: x is not one of those choices.

So do I take a dot product with a and then what else? λb or λa?

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- #8

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There's no reason to take the dot product with λa. Just use a.

That will give you two equations. You need one more. Can you find a direction that is orthogonal to both a and b?

- #9

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Ahh if you take a dot product with (axb)

- #10

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Very good.

There is one special case with which you need to concern yourself: What if axb is zero?

There is one special case with which you need to concern yourself: What if axb is zero?

- #11

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Having taken dot products with a and b, there's no point also doing λa or λb - that's just more of the same. Taking dot product with x is a bad idea because you will get scalar terms like xSo do I take a dot product with a and then what else? λb or λa?

Another approach is to take a cross-product instead of a dot product. The advantage is that you still have a vector equation, so you avoid having to reconstruct a vector equation from a bunch of scalar equations. Again, there's no point taking the cross product with x as it will result in x

- #12

- #13

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Or the same?

- #14

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I don't know? Then either a or b has to zero?

axb=0 can be zero if either or both of a and b are zero. Excluding these trivial cases, axb=0 means that a and b are either parallel or anti-parallel.Or the same?

Those trivial cases are also things to consider: What if λ=0? a=0? b=0?

You can get a general solution when none of λ, a, or b is zero, and that may be good enough for the problem at hand; check the question to see if that's all you need to cover. Those special cases may multiple solutions, a different form for the solution, or no solution at all.

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- #16

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axb=0 can be zero if either or both of a and b are zero. Excluding these trivial cases, axb=0 means that a and b are either parallel or anti-parallel.

Those trivial cases are also things to consider: What if λ=0? a=0? b=0?

You can get a general solution when none of λ, a, or b is zero, and that may be good enough for the problem at hand; check the question to see if that's all you need to cover. Those special cases may multiple solutions, a different form for the solution, or no solution at all.

So from the original

λx + (a cross x) = b

I dotted it with b, a and (axb) and got these three equations respectively

x(λb + (b cross a)) = |b|^2

λx.a = b.a

x.λ(a cross b) + (a cross b).(a cross x) = 0

How do I combine these to solve for x? Also , in the second equation is x not just equal to b/λ?

Another approach is to take a cross-product instead of a dot product. The advantage is that you still have a vector equation, so you avoid having to reconstruct a vector equation from a bunch of scalar equations. Again, there's no point taking the cross product with x as it will result in x^{2}terms. You will need the formula for expanding triple vector product: a x (b x c).

All the approaches I've looked at using the cross product make it extremely hard to isolate x no?

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- #17

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Ultimately this will give you three scalar equations for different components of x.How do I combine these to solve for x?

Note very well: Your approach is not how I would solve this problem. It's going to be a bit messy. I was trying to help you use your approach, but perhaps I should have pressed the site rules and given you a bit more help than nominal.

No! That you wrote this, coupled with the way in which you wrote your equations, suggests to me that you have a basic misunderstanding of vectors and operations on them.Also , in the second equation is x not just equal to b/λ?

It's easy to see that ##x=b/\lambda## is not a general solution because ##\lambda x + a\times x## yields ##b+(a×b)/\lambda##, and this is equal to ##b## only if ##a \times b=0##. Always double check that your solution is indeed a solution. What's happening here is that taking the inner product of each sides of ##\lambda x + a\times x = b## with ##a## loses loses information about the components normal to ##a##. What haruspex is suggesting is that you use some other operation to recover that lost information.

Hint: ##\vec u \cdot \vec v## tells you about the component of ##\vec v## that is parallel to ##\vec u##. ##\vec u \times \vec v## tells you about the component of ##\vec v## that is orthogonal the ##\vec v##. Use both and you have full knowledge of ##\vec v##.

You will need to use the triple product formulae.All the approaches I've looked at using the cross product make it extremely hard to isolate x no?

haruspex is suggesting a second approach to solving this problem. I would solve this problem by yet another approach.

First off, I would look for special cases and deal with those separately. λ=0 is an obvious special case. Setting λ to 0 fundamentally changes the nature of the problem, so handle that problem separately. a×b=0 is a less obvious special case.

If a×b≠0, then a, b, and a×b span three dimensional space, so that means you can write the solution to the problem as being of the form ##x=\alpha a + \beta b + \gamma a\times b##. Substitute this into the problem statement and solve for ##\alpha##, ##\beta##, and ##\gamma##.

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Not at all. Did you understand what two operations I'm suggesting to perform on the original equation? After using the triple vector product formula you will have three equations involvingAll the approaches I've looked at using the cross product make it extremely hard to isolate x no?

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