# Vector Scalar (dot) Product

1. Feb 1, 2014

### elevenb

I have been asked to solve the equation for x - All letters bar λ are vectors.

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. Feb 2, 2014

### Simon Bridge

Welcome to PF;

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. Feb 2, 2014

### elevenb

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

4. Feb 2, 2014

### D H

Staff Emeritus
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.

5. Feb 2, 2014

### elevenb

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

6. Feb 2, 2014

### D H

Staff Emeritus
You need more equations. That one equation alone won't do it.

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. Feb 2, 2014

### elevenb

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

Last edited: Feb 2, 2014
8. Feb 2, 2014

### D H

Staff Emeritus
Taking the dot product with λb won't add any information. All it will do is add a multiplicative factor of λ to both sides of the equality you have already developed. Divide by λ and you are back to your original equality.

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. Feb 2, 2014

### elevenb

Ahh if you take a dot product with (axb)

10. Feb 2, 2014

### D H

Staff Emeritus
Very good.

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

11. Feb 2, 2014

### haruspex

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 x2 which will be hard to remove. How could you combine a and b?

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 x2 terms. You will need the formula for expanding triple vector product: a x (b x c).

12. Feb 2, 2014

### elevenb

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

13. Feb 2, 2014

### elevenb

Or the same?

14. Feb 2, 2014

### D H

Staff Emeritus
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.

15. Feb 2, 2014

### haruspex

Pardon me for butting in again, but I really think there's a better way than generating a set of scalar equations then trying to extract a vector equation for x.
mc94, taking the dot product with a kept some information and discarded other information. Can you put into words what information was kept and what lost? What operation could you perform on the original equation that achieved the converse, i.e. kept the information the dot product with a lost and vice versa?

16. Feb 3, 2014

### elevenb

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?

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

Last edited: Feb 3, 2014
17. Feb 3, 2014

### D H

Staff Emeritus
Ultimately this will give you three scalar equations for different components of 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.

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.

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 $\gamma$.

18. Feb 3, 2014

### haruspex

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 involving x, a.x and a $\times$ x. Use two of them to eliminate the last two variables and you end with an equation for x. About 6 lines of algebra in all.