So I understand that the distance from the origin $(0,0,0)$ to a point $(x,y,z)$ is:
$$
\sqrt {x^2 + y^2 + z^2}
$$
How would I go about finding the point closest to the origin $(0,0,0)$ on the line of intersection of the planes
$$2z + y = 12$$
$$x + y = 6$$
Using the method of Lagrange multipliers. Then, calculating the distance between the origin and the point? How would I go about doing this?
I have parameterized like so, but don't know how to continue:
\begin{eqnarray}
z &=& 3+\frac{t}{2}\\
x &=& t\\
y &=& 6-t
\end{eqnarray}
Best Answer
You can instead minimize the squared distance to simplify taking derivatives since squaring is an increasing transformation (minimizing the distance is the same as minimizing the distance squared).
$$\mathcal{L} = x^2 +y^2 + z^2 + \lambda (2z+y-12) + \mu (x+y-6)$$
Then find $x,y,z$ minimizing this in terms of the lagrange multipliers. Then substitute those values back into the lagrangian above, and finally optimize the lagrange multipliers.