In class we tried to find the distance from to the plane of equation .

There are several ways of doing this. For example:

Fix a point on the plane . Any point will do, say .

Find a vector perpendicular to the . For example, .

[Again, if a plane has equation then is perpendiculart to it.

For example, is parallel to the plane , which we can rewrite as , which is the equation of the set of points perpendicular to the vector . This is to say, the plane is perpendicular to the vector . Since the plane : is parallel to , is also perpendicular to .

Another way of reaching the same conclusion is to rewrite in the form for some appropriate vector . There are many choices of and they all work; all we need is that , i.e., that is in the original plane. For example, the point belongs to the plane : , so we can rewrite the equation of as , which is equivalent to saying that But this means that the vector is perpendicular to the vector , which is an arbitrary vector in the direction of the plane.]

Let’s continue with the problem of finding the distance from to :

Consider the projection of the vector in the direction of . Clearly, the distance from to is the length of .

Recall that .

Then and the distance is .

In detail, so and , so .

Notice that (as discussed in class) the distance from a point to the line in the direction of that goes through a point is given by , while (by the above) the distance from a point to a plane containing a point and perpendicular to a vector is given by . While the expressions are similar, one involves a cross product and the other a dot product. This is because in one case we express the distance in terms of the sine of an angle , and in the other, in terms of its cosine or, what is the same, in terms of the sine of . (Drawing a diagram may help you clarify the situation.)

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3 Responses to 275- Distance from a point to a plane

The technique of almost disjoint forcing was introduced in MR0289291 (44 #6482). Jensen, R. B.; Solovay, R. M. Some applications of almost disjoint sets. In Mathematical Logic and Foundations of Set Theory (Proc. Internat. Colloq., Jerusalem, 1968), pp. 84–104, North-Holland, Amsterdam, 1970. Fix an almost disjoint family $X=(x_\alpha:\alpha

At the moment most of those decisions come from me, at least for computer science papers (those with a 68 class as primary). The practice of having proceedings and final versions of papers is not exclusive to computer science, but this is where it is most common. I've found more often than not that the journal version is significantly different from the […]

The answer is no in general. For instance, by what is essentially an argument of Sierpiński, if $(X,\Sigma,\nu)$ is a $\sigma$-finite continuous measure space, then no non-null subset of $X$ admits a $\nu\times\nu$-measurable well-ordering. The proof is almost verbatim the one here. It is consistent (assuming large cardinals) that there is an extension of Le […]

I assume by $\aleph$ you mean $\mathfrak c$, the cardinality of the continuum. You can build $D$ by transfinite recursion: Well-order the continuum in type $\mathfrak c$. At stage $\alpha$ you add a point of $A_\alpha$ to your set, and one to its complement. You can always do this because at each stage fewer than $\mathfrak c$ many points have been selected. […]

Stefan, "low" cardinalities do not change by passing from $L({\mathbb R})$ to $L({\mathbb R})[{\mathcal U}]$, so the answer to the second question is negative. More precisely: Assume determinacy in $L({\mathbb R})$. Then $2^\omega/E_0$ is a successor cardinal to ${\mathfrak c}$ (This doesn't matter, all we need is that it is strictly larger. T […]

The power of a set is its cardinality. (As opposed to its power set, which is something else.) As you noticed in the comments, Kurepa trees are supposed to have countable levels, although just saying that a tree has size and height $\omega_1$ is not enough to conclude this, so the definition you quoted is incomplete as stated. Usually the convention is that […]

The key problem in the absence of the axiom of replacement is that there may be well-ordered sets $S$ that are too large in the sense that they are longer than any ordinal. In that case, the collection of ordinals isomorphic to an initial segment of $S$ would be the class of all ordinals, which is not a set. For example, with $\omega$ denoting as usual the f […]

R. Solovay proved that the provably $\mathbf\Delta^1_2$ sets are Lebesgue measurable (and have the property of Baire). A set $A$ is provably $\mathbf\Delta^1_2$ iff there is a real $a$, a $\Sigma^1_2$ formula $\phi(x,y)$ and a $\Pi^1_2$ formula $\psi(x,y)$ such that $A=\{t\mid \phi(t,a)\}=\{t\mid\psi(t,a)\}$, and $\mathsf{ZFC}$ proves that $\phi$ and $\psi$ […]

Yes, the suggested rearrangement converges to 0. This is a particular case of a result of Martin Ohm: For $p$ and $q$ positive integers rearrange the sequence $$\left(\frac{(−1)^{n-1}} n\right)_{n\ge 1} $$ by taking the ﬁrst $p$ positive terms, then the ﬁrst $q$ negative terms, then the next $p$ positive terms, then the next $q$ negative terms, and so on. Th […]

Yes, by the incompleteness theorem. An easy argument is to enumerate the sentences in the language of arithmetic. Assign to each node $\sigma $ of the tree $2^{

Have we gotten our Homework 2 assigned yet?

Hi William,

HW 2 is posted now under syllabus.

Thanks a bunch