Nice job, yeah that is a pain in the ass to do by hand.

dam i forgot how to use mathcad, the unit shit is so annoying in the program

it's a homogeneous system. can't you just row reduce it like normal? then when you get free variables, use your x^2 + y^2 + z^2 = 1 constraint?

Just curious only taken calc 1-2 enough to pass for my major requirements.

My question is can those of you more well versed in math in general visualize what the equations mean/are saying by quickly glancing at more complex equations? I'd like to learn more about your general mindset from your perspective.

Feel free to pm me if you don't mind giving me an answer!

I would like to hear this answer as well

If you can use a TI-83, this is very simple to solve a system of equations using matrices. just input the equations into a matrix and find the rref (reduced row echelon form).
Here's a quick tutorial

Of course you can, as long as det(A) is not 0, where A is the matrix of your coefficients.

On a pure maths perspective, the system formed by the 3 first equations is linear, with a non-zero determinant.
Hence, for every column C, the equation A*(x, y, z) = C has exactly 1 solution.

In your example, the solution is obvious : it is x = y = z = 0. Notice that there are no different values for x, y, z that also satisfy the first 3 equations.

Now you're talking about a fourth equation that x, y and z must also satisfy : x²+y²+z² = 1.

Since the previous solution doesn't answer that constraint, it is impossible to find (x, y, z) satisfying the 4 equations altogether.

I wonder about the exact writing of the problem, and about what class you're following.

Sometimes, but it just depends on the equations. For example, the intuition here isn't too difficult if you think about it:

Like the first 3 equations in this system are linear -- so plugging in all the (x,y,z)'s that satisfy one of the equations draws a straight line in 3D. So those 3 equations make 3 lines. A solution to those 3 would be a point that's on all 3 lines -- i.e. where they all meet (if it exists).

Now for that last equation, remember sqrt(x^2+y^2+z^2) is the distance formula from (x,y,z) to the origin (it just comes from using the Pythagorean theorem a couple times), so taking

sqrt(x^2+y^2+z^2) = 1 (the set of all points who are 1 unit from the origin satisfy this)
x^2+y^2+z^2 = 1 (same points satisfy this -- so the set of all points 1 unit from the origin makes a sphere of radius 1 centered at the origin)

So what you're looking for here is a point where all 3 lines meet ALSO on the surface of that sphere.

feanur is right, (0,0,0) is the unique solution to the first 3 equations since A has an inverse (which immediately tells us there is no solution satisfying all 4 equations because it's the only answer for the first 3 equations, but isn't on the unit sphere).

By plugging in the point listed above -- (0.09123, -0.87391, .047745) -- we can easily see it doesn't actually satisfy the first equation.

This post was edited by Amaston on Sep 20 2015 09:29am

I'm sorry, it satisfies the first equation, but not the third.

its a negative .09123.



did that, but it has to satisfy x^2 + y^2 + z^1 = 1



this is an advance mechanics of materials course, this topic is about 3d stress.

is there anyway to program all 4 of these equations into Matlab and get a solution from matlab? Cause I need to do it 2 more times for other values