# Vector equation question.

1. May 12, 2012

### Lucy Yeats

1. The problem statement, all variables and given/known data

Four 3-vectors a, b, c, and d are related by the equation ax + by + cz = d; where x, y, and z are real parameters. Using a suitable combination of scalar and vector products, find x, y, and z in terms of the vectors.

2. Relevant equations

3. The attempt at a solution

I can solve the equations in component form, but have no idea how to solve in terms of scalar and vector products only.

2. May 12, 2012

### I like Serena

What if you multiply your equation with, say, $\mathbf{a}$?

3. May 12, 2012

### Lucy Yeats

So I'd have a.ax+b.ay+c.az=d.a, but I can't see the next step.

4. May 12, 2012

### I like Serena

You can do the same thing with the other 2 vectors.
This gives you 3 regular equations in x,y,z that you can solve.

5. May 12, 2012

### Lucy Yeats

Thanks! So I then get a long complicated formula by solving those equations by eliminating variables in the normal way?

6. May 12, 2012

### Steely Dan

This is correct.

7. May 12, 2012

### Lucy Yeats

This is 4 mark question that's meant to take 4 minutes in an exam, but there's quite a lot of algebra for that...

8. May 12, 2012

### I like Serena

Yep.

Actuallly, I would typically solve it using matrices.

Component-wise, you can write it as: Ax=d.
The regular solution is x=A-1d.
I believe this was your first approach?

Since you're apparently not supposed to use the components, the logical follow up is to do:
ATAx=ATd
This is basically, what you just did, multiplying the equation with each vector,
The solution is:
x=(ATA)-1ATd

The advantage of this method is that you do not have to know the components of the vectors, only their dot products.

Incidentally, this is also the method to solve a set of equations that is over determined (more equations than variables).
This is what you would use if a,b,c,d were n-vectors instead of 3-vectors.
Same method applies.

Last edited: May 12, 2012
9. May 12, 2012

### I like Serena

4 minutes is not much.
Perhaps too little to do the matrix inverse.

Perhaps you can get away by simply setting up the set of equations and say x,y,z is the solution of the set of equations, without actually doing the algebraic solution?
Or perhaps just use Cramer's[/PLAIN] [Broken] rule?

Last edited by a moderator: May 6, 2017
10. May 12, 2012

### Steely Dan

Well it's not an incredible amount of algebra, it's only three equations :)

11. May 12, 2012

### Lucy Yeats

So is there any way to put that formula back into vector form?

I know the elements of the matrix A^TA are all dot products, but not how to get rid of the matrix.

12. May 12, 2012

### I like Serena

You get a 3x3 matrix with dot products.
A matrix is just a handy shorthand notation to write it down.

You can work it out without matrix, or eliminate the matrix afterward.
But that will give you a set of 3 equations that is rather long to write down.

13. May 12, 2012

### Lucy Yeats

How will I get rid of the components of A^T when I multiply out that expression?

14. May 12, 2012

### I like Serena

I'm not sure what you mean.

Let's see:

AT d = (a.d, b.d, c.d)

ATA = ([a.a, a.b, a.c], [b.a, b.b, b.c], [c.a, c.b, c.c])

As you can see, this vector and matrix contain only the dot products.
So what you have, is a regular system of equations:
Mx=y​
with M=ATA and y=AT d.

You can solve it any way you like (matrix-inverse, Cramer's rule, or regular equation solving with substitution).
How do you mean that you need to get rid of the components of A^T?

15. May 12, 2012

### Lucy Yeats

Ah, that makes sense now. Thank you very very much! :-)