TITLE: Where is the point that has the shortest distance to the orgin?
QUESTION [0 upvotes]: I just finished an exam, it has the following question: Where is the point on the plane $3x + 5y + z = 18$ has the shortest distance to $(0,0,0)$?
I found this question similar: Find the point on the plane $2x - y + 2z = 20$ nearest the origin
But I am not following the steps:

Obtain the normal vector of the plane <3, 5, 1>
Find the Unit normal vector by dividing $\sqrt{3^2 + 5^2 + 1^2}$
Then how do I go about minimizing the problem? (Set up the constrain)?
Do I use lagrange multipliers? (We were learning about this topic).

REPLY [1 votes]: If your class was about calculus and you were learning about Lagrange multipliers you're supposed tu use it. So:
Distance from one point $(x,y,z)$ to $(0,0,0)$ is
$$
\lVert (x,y,z)-(0,0,0) \rVert = \sqrt{x^2+y^2+z^2}
$$
You want to minimize it subject to $3x+5y+z=18$. But this is the same as minimizing
$$
f(x,y,z) = x^2+y^2+z^2
$$
subject to $3x+5y+z=18$. Then we say $g(x,y,z)=3x+5y+8z$.
$$\nabla f(x,y,z) = (2x,2y,2z) \quad\text{and}\quad \nabla g(x,y,z)=(3,5,8)$$
Then we have to solve $\nabla f(x,y,z) = \lambda\nabla g(x,y,z)$ with the same subject. That's the system of equations:
$$\left.
\begin{array}{c}
2x = 3\lambda \\
2y = 5\lambda \\
2z = 8\lambda \\
3x+5y+8z = 18 \\
\end{array} \right\}
$$
After this, you have to prove if the point you find is actually a minimum.
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