Two subspaces of different dimensions and the orthogonal complement

I need help getting started with this problem. Please just give me a nudge in the right direction.

Suppose $U_1,U_2$ are subspaces of the euclidean space $V$ such that $dimU_1.
Show that there is a nonzero vector $uin U_2$ such that $u$ is in the orthogonal complement of $U_1$.

What are the exact dimensions of a DUPLO brick?

The dimensions of LEGO brick are documented well, but what about DUPLO bricks? If have only found the general statements that DUPLO is twice as large as LEGO but no exact numbers. What about the thickness of walls and the space that must be subtracted at each side (0.2mm for LEGO), is it also just doubled?. If you compare the bottom of a LEGO 2×4 brick and a DUPLO 2×4 brick, there are also additional small walls that I have not found documented.


Update: Some dimensions of a DUPLO brick can be derived from the dimensions of a LEGO brick, which have been documented by Robert Cailliau among
others:

Above and below a DUPLO 2×2 brick one can put two LEGO 2×4 bricks (or four 2×2 LEGO bricks). The outer walls of such stack seem to align perfectly, so the width and length of a DUPLO 2×2 brick is equal to the width of a LEGO 2×4 brick, which is 31.8mm (32mm-0.2mm). By putting DUPLO and LEGO side by side, one can also see that the DUPLO height without studs is twice of the size in LEGO (19.2mm). The LEGO studs below the DUPLO brick touch DUPLO walls, so DUPLO wall thickness is the same as LEGO wall thickness (either before spline walls were introduced to LEGO bricks or LEGO wall thickness plus the length of LEGO splines), which is 1.5mm. The DUPLO tube height is the height of the brick minus the height of a LEGO stud (which is somewhere between 1.6mm and 1.8mm). I’d better use the larger value, so the tubes start 1.8mm above the floor of the brick. From the top, LEGO tubes fit into DUPLO studs, so the stud inner diameter should equal to 6.51mm (8*√2 – 4.8$). But if you carefully look inside a DUPLO stud, this fitting is only achieved by four additional lugs that give a minimal inner diameter around 6mm to facilitate snapping into.

The remaining dimension include:

  • stud outer diameter
  • stud height
  • tube inner diameter (but DUPLO tubes are not fully circular inside
    on a closer look!)
  • tube outer diameter (also occurs at some other elements for instance as stem at #6510)
  • spline length

Spline length can be derived from wall width and stud outer diameter. Tube
outer diameter and stud outer diameter can also be computed from each other. So basically three dimensions are left to be explained. Comments and insights are welcome how more of the remaning dimensions can be
derived from standard LEGO/DUPLO dimensions.

Does Coulomb’s Law, with Gauss’s Law, imply the existence of only three spatial dimensions?

Coulomb’s Law states that the fall-off of the strength of the electrostatic force is inversely proportional to the distance squared of the charges.

Gauss’s law implies that a the total flux through a surface completely enclosing a charge is proportional to the total amount of charge.

If we imagine a two-dimensional world of people who knew Gauss’s law, they would imagine a surface completely enclosing a charge as a flat circle around the charge. Integrating the flux, they would find that the electrostatic force should be inversely proportional to the distance of the charges, if Gauss’s law were true in a two-dimensional world.

However, if they observed a $frac{1}{r^2}$ fall-off, this implies a two-dimensional world is not all there is.

Is this argument correct? Does the $frac{1}{r^2}$ fall-off imply that there are only three spatial dimensions we live in?

I want to make sure this is right before I tell this to my friends and they laugh at me.

Are all bivectors in three dimensions simple?

I want to show that all bivectors in three dimensions are simple.

If I understand correctly, a bivector is simply an element from the two-fold exterior product $bigwedge^2V$ of a vector space $V$, right?

Now consider $V=mathbb{R}^3$ with the standard base $(e_i)_{ileq3}$ and the bivector $v=e_1wedge e_2+e_1wedge e_3+e_2wedge e_3=e_1wedge(e_2+e_3)+e_2wedge e_3$.

Clearly, this is not simple! I guess I have some sort of misconception. Can someone explain to me what I did wrong?

What exact dimensions does a physical cone AoE template need to have?

I am trying to get a precise set of measurements for cone-shaped areas of effect so that I can craft a physical template to use as an overlay when playing with miniatures. I want to get it right the first time.

The problem is that the rules for the shape of a cone AoE aren’t very precise, and when you start looking at them closely, there are a bunch of inconsistencies. These are “good enough” when playing without a grid, or when only needing a grid-shaped approximation of the area of effect, but they are causing me difficulties with crafting an accurate physical template.

There are three incompatible interpretations of the maximum length of a cone (PHB, p. 204; emphasis mine):

A cone’s area of effect specifies its maximum length.

… which each lead to slightly different template constructions (using a 30′ length as the working example):

  1. You can use the maximum length as the centreline of an isosceles triangle.

    This would give a template that has a centreline 30′ long, but edges beyond the “maximum” length that are ~33.5′ long. The end of the area of effect would be a flat plane. This shape would encourage players to aim off-centre to use the “bonus” length offered by the edges.

  2. You can use the maximum length as the edges of an isosceles triangle.

    This would give a template with no portion farther than the maximum away, but with a centreline length that is only ~27.8′. The end of the area of effect would be a flat plane. This shape would motivate players to do off-centre aiming in order to get full use of the maximum-length edges of the template.

  3. You can use the maximum length as a radius of a conical section of a sphere (a spherical cone).

    This would result in a template with a rounded end, with all points on that arc exactly the maximum length away from the point of origin.

Which of these interpretations of the definition of a cone’s area of effect is the correct one to base my template design and construction on?

What are the exact dimensions of a DUPLO brick?

The dimensions of LEGO brick are documented well, but what about DUPLO bricks? If have only found the general statements that DUPLO is twice as large as LEGO but no exact numbers. What about the thickness of walls and the space that must be subtracted at each side (0.2mm for LEGO), is it also just doubled?. If you compare the bottom of a LEGO 2×4 brick and a DUPLO 2×4 brick, there are also additional small walls that I have not found documented.


Update: Some dimensions of a DUPLO brick can be derived from the dimensions of a LEGO brick, which have been documented by Robert Cailliau among
others:

Above and below a DUPLO 2×2 brick one can put two LEGO 2×4 bricks (or four 2×2 LEGO bricks). The outer walls of such stack seem to align perfectly, so the width and length of a DUPLO 2×2 brick is equal to the width of a LEGO 2×4 brick, which is 31.8mm (32mm-0.2mm). By putting DUPLO and LEGO side by side, one can also see that the DUPLO height without studs is twice of the size in LEGO (19.2mm). The LEGO studs below the DUPLO brick touch DUPLO walls, so DUPLO wall thickness is the same as LEGO wall thickness (either before spline walls were introduced to LEGO bricks or LEGO wall thickness plus the length of LEGO splines), which is 1.5mm. The DUPLO tube height is the height of the brick minus the height of a LEGO stud (which is somewhere between 1.6mm and 1.8mm). I’d better use the larger value, so the tubes start 1.8mm above the floor of the brick. From the top, LEGO tubes fit into DUPLO studs, so the stud inner diameter should equal to 6.51mm (8*√2 – 4.8$). But if you carefully look inside a DUPLO stud, this fitting is only achieved by four additional lugs that give a minimal inner diameter around 6mm to facilitate snapping into.

The remaining dimension include:

  • stud outer diameter
  • stud height
  • tube inner diameter (but DUPLO tubes are not fully circular inside
    on a closer look!)
  • tube outer diameter (also occurs at some other elements for instance as stem at #6510)
  • spline length

Spline length can be derived from wall width and stud outer diameter. Tube
outer diameter and stud outer diameter can also be computed from each other. So basically three dimensions are left to be explained. Comments and insights are welcome how more of the remaning dimensions can be
derived from standard LEGO/DUPLO dimensions.

Product of two multivariate gaussian distributions of different dimensions

I have to multiply $mathcal{N}(mu_1,Sigma_1)$ with $mathcal{N}(mu_2,Sigma_2)$, where $mu_1 in mathbb{R}^{text{Dx1}}$ and $mu_2 in mathbb{R}^{text{1xM}}$.
I’m thinking that the result might be a matrix normal distribution like
$$mathcal{MN}(mu_1mu_2,Sigma_1, Sigma_2)$$
Is this correct?