NEED YOUR HELP. MID-TERM EXAM IN 6 HOURS.?

Question

Guys! Huhu. I need your help! please help me on this.. please please. I'm pleading already..

My teacher gave us a hint that there's an item on the exam that is like this:

Find the point on the plane given by x + y + 2z = 2 which is closest to the origin. DO NOT USE LAGRANGE.

Wah. I do know how to solve it with lagrange multipliers, but I dont have any idea how i would solve it in other way.

seems like, given

x + y + 2z = 2

solve for z = (2 -x - y)/2

then sub the value of z to distance formula;

D = sqrt(x^2 + y^2 +z^2), then get the first derivative of D in terms of x and y, then equate to zero, then solve for values of x and y, then z.

the final answers are x = 1/3, y=1/3, z=2/3, and im not getting these answers. Im really troubled.

Please help me out. I'd really appreciate it. Please please. My exam is in 6 hours...

Answer 1

I dint read the whole thing but x + y + 2z = 2 is probly the answer and she accidently said it.

Answer 2

You are correct in your strategry.

Take the derivative of D with respect to x and then with respect to y and set it equal to zero.  I get x = 1/3, y = 1/3, therefore z = 2/3.

I would show all the steps, but it's a bit complicated in this limited text box.

One suggestion I might make is to NOT take the derivative of D, but of D^2, that way you don't need to worry about the square root.  If you minimize D or D^2 you will get the same x and y values.

Take care,

David

Answer 3

If we want to minimize the distance, that is the same thing as minimizing the distance squared (minimum D is also a minimum D^2).  This will make the derivative easier, so just let E = D^2

E = x^2 + y^2 + z^2 = x^2 + y^2 + (1/4)*(2-x-y)^2

the following derivatives are all partials with respect to the denominators:

dE/dx = 2x + (1/2)(2 - x - y)(-1) = 0

5x + y = 2

dE/dy = 2y + (1/2)(2 - x - y)(-1) = 0

x + 5y = 2

solving the system of equations gives x = 1/3, y = 1/3

Plugging those in for z = (2-x-y)/2 gives z = 2/3

In case you made the same error I originally made, I got (1/2, 1/2, 1/2) for my first answer.  The error I made was writing z as:

z = [(2-x-y)/2]^2

That part is correct, but when I did the derivative, I let the exponent of two cancel out the denominator of 2 completely, instead of realizing the denominator is actually 2^2, and after cancelling with the exponent, a 1/2 factor should actually be left.

Answer 4

I believe the simplest way to solve this problem is to recognize that a normal vector to the surface of this plane at any point is equal to the coefficients of x, y, and z  (1,1,2).  

You need to find a point on the plane where a vector pointing from the origin to that point is exactly parallel to the normal vector that defines the plane.  

The easiest way to do this is to define a vector "v" such that the cross product of v and the normal "n" is equal to 0.  

Let's work it out...

n X v = 0

(z - 2y)i + (2x - z)j + (y -x)k = 0

For this to be true, all 3 terms in parenthesis have to equal 0.

Use 2 of these equations and the equation of the plane to solve for x, y and z.  

I used y = x, z = 2y, and x + y + 2z = 2 as my equations.

From these, it is fairly simple to see the only possible value for x, y, and z are (1/3, 1/3, 2/3).  At this point on the plane, a line can be drawn perpendicularly to the origin.  This is the shortest line that can be drawn from the plane to the origin and so it is the closest point on the plane to the origin.