How to Solve a Vector Equation Using Scalar and Vector Products?

[Lucy Yeats]

Homework Statement

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.

Homework Equations

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.

 

[I like Serena]

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

 

[Lucy Yeats]

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

 

[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.

 

[Lucy Yeats]

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

 

[Steely Dan]

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

This is correct.

 

[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...

 

[I like Serena]

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

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:
A^TAx=A^Td
This is basically, what you just did, multiplying the equation with each vector,
The solution is:
x=(A^TA)^-1A^Td

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.

 

[I like Serena]

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...

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] rule?

 

[Steely Dan]

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...

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

 

[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.

 

[I like Serena]

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.

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.

 

[Lucy Yeats]

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

 

[I like Serena]

I'm not sure what you mean.Let's see:

A^T d = (a.d, b.d, c.d)

A^TA = ([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=A^TA and y=A^T 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?

 

[Lucy Yeats]

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