Closest point to a set with l1 norm

[CCMarie]

Homework Statement

Given the plane (π):x+2y+z-1=0, what is the element closest to the origin when the distance is measured with l1 norm.

Relevant Equations

The l1 norm for a vector t=(x_0,y_0,z_0) is
||t|| = |x_0|+|y_0|+|z_0|

I tried to find the element of best approximation
||t_0||≤||t||, ∀ y ∈ π

Then |x_0|+|y_0|+|z_0| ≤|x|+|y|+|z| and we have x_0+2y_0+z=1 and x+2y+z=1.

But I don't know hoe to continue...

 

[haruspex]

The only rigorous way I know to solve this would be to consider the various combinations of signs the coordinates could have. For each, the metric can be treated as a linear function. In principle there would be eight cases, but there is a symmetry which allows you to whittle that down.

 

[haruspex]

Being a little less rigorous, from visualising the plane, it looks very likely all coordinates will be positive.

 

[CCMarie]

Thank you!