What is the closest point on a plane to the origin using vectorial reasoning?

[Microdude]

As per Astronuc request, i will post this here so it can benifiet other people who may have similar questions.

I have this Linear Algebra assignment to do and i was completely stomped on how to go about to answer this question. I asked Astronuc for help to clarify a bit what was asked to understand it more.

Homework Statement

the question goes as follows,
If a>0,b>0,c>0. find the point on the plane x/a+y/b+z/c=1 which is closest to the origin (0,0,0)

i need to justify this using vectorial reasoning.

2. The attempt at a solution
after sending a PM to Astronuc i came across a site which described me something related to my problem. unfortunately i cannot find it anymore i printed out what was written there.

Astronuc replied to me with the following suggestions.

The shortest distance between two points is a straight line. The linear combination of two vectors defines a plane, and the cross product of those two vectors is perpendicular to that plane. So find the vector which is perpendicular to the plane and passes through the origin.

 

[Dick]

By that approach, your first step is to find a vector perpendicular to the plane. Are you up for that? Another approach is simply to say you want to minimize the distance between (0,0,0) and (x,y,z) subject to the constraint that (x,y,z) satisfies your plane equation. Eliminate a variable and set partial derivatives wrt to the two remaining variables equal to zero. Does that sound like something you've done before?

 

[Microdude]

hmm.. not really.. but i think what is asked is to find the vector perpendicular to the plane. i'll try it that way. cause it must be done using vector reasoning's. I'm trying it out now! thanks!

 

[Dick]

If you want to do it using vectorial methods, then your plane equation has the form f(x,y,z)=c. Think about the gradient vector of f. PS if you are trying it via minimization, minimize the square of the distance rather than the distance. Makes life easier.