Question

: Calculus/Linear Algebra: Find the point on the plane x+y-z=4 which is closest to the point (2,1,0).

Answer 1

(2,2,0) is one y value apart from it and it satisfies:
2+2-0=4
4=4

(2,1,-1) is one z value apart from it and satisfies:
2+2+-1=4
4=4

(3,1,0) is one x value apart and it satisfies:
3+1+0=4
4=4

Answer 2

I'm sorry, I'm don't understand how (2,2,0) is one y and etc. value when we're looking for x, y, z as coordinates i.e. x, y, z are single values. Your Y(2,2,0) and such are themselves points on the plane, but i'm looking for that single point that is closest to (2,1,0)

Answer 3

What I was showing was the direction of the movement, there are 3 points (above) that are equidistant from the point you provided.

The first is a difference of 1 unit in the y direction, the 2nd a 1 unit difference in the z, and the last 1 unit in x.

Answer 4

I suppose we could do some distance formulae to make sure they are equidistant, I just kinda assumed they were.

Answer 5

\[\sqrt{(2-2)^2+(2-1)^2+(0-0)^2}=1\]
So, movement in the y is, in fact 1.

\[\sqrt{(2-2)^2+(1-1)^2+(-1-0)^2}=1\]
So, movement in the z is, in fact 1.

\[\sqrt{(3-2)^2+(1-1)^2+(0-0)^2}=1\]
So, movement in the x is, in fact 1.

Answer 6

So, You can assume your answer...

Take: F(x,y,z) = x+y-z
Find the closest point to (2,1,0) on the plane F(4):

(show distance formulae, defining them as r(x,y,z))
\[\large\therefore \exists 3 (x,y,z)|\Delta r(x,y,z)=1\]
(list the points)
QED

Answer 7

Should be a comma after 3
I suck at typing math shorthand >.<

Answer 8

Also, you can expand your proof by switching to vectors... define the distance formulae as the vector r and then you can show that the movement is a unit vector in each case:
\[\Delta r_x=i\]\[\Delta r_y=j\]\[\Delta r_x=k\]
All of those should have "hats" denoting the unit vectors.