Expert: Paul Klarreich - 2/21/2011

Question
Hi!

I'm trying to find the point (x,y,z) on the plane 2x-y+z=3 that is closest to the point (2,0,-1).

I used method of elimination of variables, and got the point to be (-5/7 , 18/7, 13/7)

where:
x = -5/7
y = 18/7
z = 13/7

Is this the correct answer?  Also, would a question like this be easier solved via Lagrange Multipliers?

Answer
Questioner: mike
Country: United States
Category: Calculus
Private: No
Subject: Max/Min question
Question: Hi!

I'm trying to find the point (x,y,z) on the plane 2x-y+z=3 that is closest to the point (2,0,-1).

Subbing:

2(2) - 0 - 1 = 3
4 - 1 = 3

THIS POINT IS IN THE PLANE!!

This point itself is the answer!

(see below)
.....................


I used method of elimination of variables, and got the point to be (-5/7 , 18/7, 13/7)

where:
x = -5/7
y = 18/7
z = 13/7

Is this the correct answer?  Also, would a question like this be easier solved via Lagrange Multipliers?
.........................................

The normal vector to 2x - y + z = 3  is <2,-1,1>

The line that is normal to 2x - y + z = 3 and passes through the point (2,0,-1) has parametric equations:

x - 2 = 2t
y - 0 = - t
z + 1 = t

OR:

x     = 2t + 2
y     = - t
z     = t - 1

Substitute into the plane equation:

2(2t + 2) - (-t) + (t - 1) = 3
4t + 4 +  t - 1 = 3

5t + 3  = 3

t = 0  What?  t = 0??

That means this is the point itself.  (See above.)