How can I relate the idea of having the minimum value of a function defined by x,y and z and planes and matrices A function is given as f(x,y,z)=2x^(2)+y^(2)+5z^(2) then the question is to find the  minimum value of this function on the plane x+2y+3z=1
I can see that the given function has always non negative value and so the minimum value is 0 but since the point (0,0,0) is not on the plane, I took g(x,y,z)=x+2y+3z and I calculated the value of f when g=1 and ∇f=λ∇g that is,
(4x,2y,10z)=λ(1,2,3) with this I found that the function has its minimum value on the plane at the coordinate (5/63,20/63,6/63) and that corresponds to a functional value of 10/63.
Is there any other way of getting the solution using matrices or the concept of planes and 3d vectors? This question was at the end of a chapter on 3d vectors and matrices in calculus 1 and so I think I am not supposed to use Lagrange multipliers. I would be grateful.
 A: That looks like a pretty standard "Lagrange Multiplier" problem.  At a max or min of f(x,y,z) subject to constraint g(x,y,z)= constant, the gradients of f and g will be parallel: $\nabla f= \lambda \nabla g$ for some constant, $\lambda$ (the "Lagrange Multiplier").
Here, $f(x,y,z)= 2x^2+ y^2+ 3z^2$ so $\nabla f= 4x\vec{i}+ 2y\vec{j}+ 10z\vec{k}$ and $g(x,y,z)= x+ 2y+ 3z$ so $\nabla g= \vec{i}+ 2\vec{j}+ 3\vec{k}$.
We must have $4x\vec{i}+ 2y\vec{j}+ 10z\vec{k}= \lambda(\vec{i}+ 2\vec{j}+ 3\vec{k})$ so we must have $4x= \lambda$, $2y= 2\lambda$, and $10z= 3\lambda$.  Since a value for $\lambda$ is not necessary for a solution to this kind of problem, often the best way to solve these equations is to divide one equation by the other.
For example, if we divide the first equation by the second, we get $\frac{4x}{2y}= \frac{1}{2}$ and then have y= 4x.  If we divide the first equation by the third, $\frac{4x}{10z}= \frac{1}{3}$ and then z= (6/5)x.  Then the constraint becomes x+ 2y+ 3z= x+ 8x+ 18/5x= (63/5)x= 1 so x= 5/63.  Then y= 4(5/63)= 20/63 and z= (6/5)(5/63)= 6/63= 2/21.   
Check my arithmetic!
