Does Arcane Eye transmit information if the caster moves to a different plane?

The description of arcane eye states that the eye created by the spell “can’t enter another plane of existence.” PHB p. 214.

Consider this scenario: a warlock casts arcane eye (perhaps using the Visions of Distant Realms invocation), creating an eye 30 feet away; subsequently casts demiplane; and then steps through the shadowy door created by demiplane, leaving behind the eye exactly where it was created.

What result? Is there any rules-as-written reason to rule otherwise than that the eye did not “enter” another plane, and so remains in existence and continues to transmit information across the planes? Or does the eye remain in existence but cease to transmit information? Or does it cease to exist? Or something else?

Orthogonal Projection of vector onto plane

I’m currently trying to learn Mathematica, and I’ve got some linear algebra tasks to solve with it. I’ve gotten quite far but now I’m stuck on this one exercise. The instructions are:

With the help of Mathematica-commands, draw a new picture, where you can see the orthogonal projection of the vector onto the plane. It should look something like this: enter image description here

Now, I started out by drawing the vector in the 3D plane with this code:

Graphics3D [ { Thick , Arrow [ { { 0 , 0 , 0} , { 1 , −1 , 2 } } ] , InfinitePlane [ { { 1 , 0 , 0} , { 1 , 1 , 1} , { 0 , 0 , 1 } } ] } , Axes −> True , AxesLabel −> { "X" , "Y" , "Z" } ] 

This gave me the 3D image in the picture above, without the projection (the dashed line) obviously. But now I’m stuck, and my question is, how would I get the orthogonal projection of the vector?

Thanks in advance.

Can I get cheaper plane tickets by using a VPN?

Use NordVPN to get cheaper tickets on KAYAK?

In a Youtube video by ADVChina, the speaker is recommending to use a VPN in China for various reasons, and to sign up using his referral code.

He says there are various practical uses including “You could also go and get cheaper plane tickets by hopping around different countries.” and shows a screenshot of a flight website called KAYAK, with the VPN software changing his country to Italy.

Is it possible to get cheaper plane tickets by using a VPN in this way? I’ve never heard of KAYAK, but if I use something like Skyscanner, is it common for tickets to be priced differently depending on which country the user’s IP is from?

Is it because of different taxes in different countries? So could it be illegal to get discounts like this? Or is there some pricing strategy from airlines to charge different amounts depending on the country of purchase?

On what plane can the keep acquired via the Throne card be on?

From the Deck of Many Things (DMG, p. 164):

Throne. You gain proficiency in the Persuasion skill, and you double your proficiency bonus on checks made with that skill. In addition, you gain rightful ownership of a small keep somewhere in the world. However, the keep is currently in the hands of monsters, which you must clear out before you can claim the keep as yours.

It says “somewhere in the world“; does this mean on the Material Plane, or could it potentially be on any plane of existence? I know it’s left intentionally vague to allow the DM more flexibility, but I wondered if “the world” at least implies the Material Plane…

Point rank in 2D plane time complexity?

I’m reading about the algorithm of finding the ranks of all points in a 2D plane, I don’t understand the time complexity formula for it. It has four steps:

  1. Compute the median of x-coordinates of all point, and divide the plane into two half Left, and Right.
  2. Recursively do 1. then when there is only 1 point, rank(that point)=0.
  3. Sort points by y-coordinate in Left, and Right separately.
  4. Update Right.

I understand the idea of these steps, and 3. has complexity $ O(n\log n)$ , but the time complexity formula in my book is

$ $ T(n)=2T(n/2)+\Theta(n),$ $

why the last term is not $ \Theta(n\lg n)$ ? That is my current idea is that the $ T(n)=\Theta(n\lg^2n)$ , by applying master’s theorem.

If the focal plane is curved, should the outer AF points work correctly or front-focus?

Many (all?) lenses have a (slightly) curved focal plane. In general it’s close enough to flat that you don’t really notice it, but under certain conditions it may become apparent. As I understand it the curved focal plane is most likely to cause issues with wider and/or wide-aperture lenses (e.g. LensRentals on the 50/1.2).

I’ve seen various threads (e.g. here, here, and here) with people having trouble getting such lenses to focus correctly using the outer AF points, yet the centre AF point works fine. Seems to come up with lenses like the Sigma 30mm f/1.4 on APS-C, or Canon 50mm f/1.2 on FF. Many responders put it down to the lenses having a curved focal plane, but my basic understanding of the way AF works makes this kind of hard to believe.

Is it normal to be unable to use the outer AF points on a normal to wide angle lens at very large apertures (~f/1.4)? Or should all AF points get the focus right for that point, no matter the curvature of the focal plane?

It seems to me that unless the AF points are trying to do something ‘smart’, using the known curvature of the focal plane, for example, they should all only provide good contrast at their specific location on the frame. So I can’t really see how the curvature of the focal plane should affect focus achieved from an AF point provided the shot isn’t recomposed after focusing.

To complicate the various discussions on the matter, some people use poor techniques, like testing in poor light, small or low contrast targets, without a tripod, etc. This makes many of the forum threads on this very hard to follow, and makes me doubt what everyone is saying (on both sides). However, there’s plenty of cases of people using sensible targets (large, 2D regions with vertical lines, daylight, tripod, no recompose) and finding their camera+lens front focuses consistently with the outer AF point but is consistently correct with the centre AF point(s). Seems to happen only/mostly on these normal-wide, large aperture lenses. So I’m wondering if it’s a normal thing, or if it’s just a lens alignment/calibration/tolerance issue that shows up more easily with these lenses?

I just bought a Sigma 30mm f/1.4 (the new Art one) for my 7D and have experienced exactly this problem. My first tests were using the centre AF point up close and it worked perfectly. I assumed I had a good copy and started shooting. Went to shoot a subject from about 6m away using the far-left AF point and could not get it to focus correctly; the focal plane was over a metre in front of the subject. So I thoroughly tested it with a well lit high contrast focus target (black & white printed A4) and am finding I need to apply autofocus micro-adjustments of about +4 using the centre AF point, +8 for many of the outer AF points, and +12 on the far left AF point (this is at about 2-6 metres away, it’s less dramatic at 30cm, where +0 is fine for all AF points with the target within the 5mm DoF).

My other lenses (including 50 f/1.8) don’t exhibit noticeable AF differences with different AF points, so I’m fairly sure it’s not a problem with my AF sensor.

And it’s not a pixel-peeping problem, it was clear from just looking at the 3″ photo on the back of the camera that it’s not focusing properly (DoF is about 1m wide, and it was front-focused by about 1m).

If it was vaguely consistent (+/- 3 say) across the whole frame & across all focal distances, I’d just pick an adjustment and get on with it. But no matter what MFA setting I pick, some AF points will be unusable. Even if it was just consistent across the frame, but varied with focal distance, I’d get the Sigma USB dock and recalibrate, but that won’t fix AF variation across the plane, only for varied focus distances.

So is this normal? Is it reasonable to expect that I can only reliably use the centre AF point on a normal-wide, large aperture lens? Or have I (and others) got reason to return lenses performing this way, expecting usable AF across the whole frame?

Update: Sent the vendor AF test shots, they claim its just normal AF variation, like @MichaelClark suggests may be the case. Could be I need to calibrate it with Sigma’s dock and/or learn to AF.

Does Hurl Through Hell dispel effects that require the target to be in the same plane as the caster?

The exact circumstances of my situation is that a vampire has charmed my fighter buddy and i wish to use my ability to make the vampire briefly go to lower planes resulting in the charm being broken though i am not sure if the hell part is supposed to be merely fluff with no actual plane travel happening. So does Hurl Through Hell cancel a vampires charm effect?

Finite subgroups of motoins always fix some point on the plane.

Let $ G$ be a finite subgroup of the group of motoins $ M$ on the plane. Then there exists a point on the plane which is left fixed by every element of $ G$ .

The proof of this is sketched by our instructor as follows $ :$

Let $ s$ be an element on the plane. Let $ S$ be the set of all images of $ s$ under the action of the elements of $ G$ on $ s$ . Then every element of $ G$ permutes the elements of $ S$ . To see this let us take $ f \in G,s’ \in S$ . If we can show that $ f(s’) \in S$ then we are through. Since $ s’ \in S$ $ \exists$ $ g \in G$ such that $ g(s) = s’$ . Since $ f,g \in G$ so $ fg \in G$ . Therefore $ fg (s) \in G$ i.e. $ f(s’) \in S,$ as claimed. Since $ G$ is finite, $ S$ is also finite. Let $ S = \{s_1,s_2, \cdots , s_n \}.$

After that our instructor asserted that the element $ s^{*} = \frac {1} {n} (s_1 + s_2 + \cdots + s_n)$ is fixed by every element of $ G$ . This is stage where I am struggling. Why $ s^{*}$ is fixed by every element of $ G$ . If the element is rotation about origin or reflection about a line passing through origin then I have understood that the claim is true because they are linear operators which act linearly on $ s^{*}$ and permute the $ s_i$ ‘s. Since vector addition is commutative we get $ s^{*}$ back by acting them on $ s^{*}$ . Also clearly $ G$ doesn’t contain any proper translation since they are of infinite order. So what are the elements other than rotation about origin and translation about a line through origin? I know that the elements of $ M$ are either of the form $ t_{a} \rho_{\theta}$ (which are precisely rotations about some point fixed by it ) or of the form $ t_{a} \rho_{\theta}r$ (which are precisely glide reflections). So according to me the elements of $ G$ is either of the form $ t_{a} \rho_{\theta}$ or of the form $ \rho_{\theta} r$ . Do they all fix $ s^{*}$ ? Please help me in this regard.

Thank you very much.