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From: Tim Little on 5 Jul 2010 01:41
On 2010-07-04, Fred Nurk <albert.xtheunknown0(a)gmail.com> wrote:
> Solve the equations 2x + 2y - z = 1 and x - y + z = 0 for x, y and z in
> terms of a parameter lamda.
>
> What does 'for x, y and z in terms of a parameter lamda' mean?

You'll find that there is more than one solution to the equations. It
turns out for these equations that there are infinitely many, all
lying along a line in 3D (x,y,z) space.

So you can use "lambda" to describe the position along that line.
There are many different ways to do that, and you're free to pick one
and give a formula for the x, y, and z coordinates on the line when
given lambda.

For linear equations you can just let lambda = one of the variables.
Then ask yourself "if we fixed a value for lambda, what would be the
solution in x,y,z to the equations?" E.g. if we let x = lambda:

2 lambda + 2y - z = 1
lambda - y + z = 0

rearrange to

2y - z = 1 - 2 lambda
-y + z = -lambda

which can be reduced to

y = 1 - 3 lambda
z = 1 - 4 lambda.

So that gives one valid solution with lambda as a parameter:

x = lambda,
y = 1 - 3 lambda,
z = 1 - 4 lambda.

There are many more. You could even pick something bizarre like
x + y = tan(lambda)^3 which would still result in a valid solution,
but be needlessly complicated.


It is possible in some systems of equations to get a different type of
solution: you end up with lambda = constant. That just means the line
is perpendicular to the axis you chose, and so that variable can't be
used to distinguish points along the line. Try a different variable,
because a line can't be perpendicular to all of the axes at once.


- Tim
From: Fred Nurk on 5 Jul 2010 02:53
Tim Little wrote:
> <snip>
> You'll find that there is more than one solution to the equations.
> <snip>
>
> <snip> You're free to pick one and give a formula for the x, y, and z
> coordinates on the line when given lambda.
>
> For linear equations you can just let lambda = one of the variables.
Then ask
> yourself "if we fixed a value for lambda, what would be the solution in
x,y,z
> to the equations?"
>
> <snip>
> It is possible in some systems of equations to get a different type of
> solution: you end up with lambda = constant. That just means the line
is
> perpendicular to the axis you chose, and so that variable can't be used
to
> distinguish points along the line. Try a different variable, because a
line >
> can't be perpendicular to all of the axes at once. <snip>

Great answer! Thanks.
From: OwlHoot on 5 Jul 2010 06:20
On Jul 5, 5:09 am, Fred Nurk <albert.xtheunkno...(a)gmail.com> wrote:
> Fred Nurk wrote:
> > <snip>
> > What does 'for x, y and z in terms of a parameter lamda' mean?
>
> The answer is:
> x = (1 - lamda) / 4, y = (3lamda + 1) / 4, z = lamda.
>
> My reasoning:
> Take z to be lamda and write equations with x and y as subject. All that
> was needed was to rearrange the second and first equations in that order.
>
> What do you think of my reasoning?

Yup, that works fine for this example.

However, it wouldn't work for any problem, such as if your equations
had been 2x + 2y - z = 1 and x + y + z = 0 (sign change of y in 2nd)

So if, say, you were writing a program to automate this for any
similar system then a slightly more general approach would be
needed, such as the one I summarized, to avoid the program
throwing wobblers every now and then for certain cases!

Cheers

John Ramsden




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