How Can We Visualize the Unit Cube in \( \mathbb{R}^4 \)?

[shwin]

I'm having trouble visualizing \ R^{4}[/itex](a domain of reals in four dimensions).<br /> <br /> 1. Describe a procedure in given 3 vectors, finds a fourth vector perpendicular to those three. Explain why we can use it in analogous fashion to the normal vector to a plane in \ R^{3}[/itex].&lt;br /&gt; &lt;br /&gt; Here, I&amp;#039;m thinking taking the normal vector of each vector and then adding the three normals would be sufficient, but I am not sure how this is analogous to the normal vector to a plane in the xyz system.&lt;br /&gt; &lt;br /&gt; 2. How many vertices does the unit cube have in \ R^{n}[/itex] have? What is the furthest distance from the origin that one can be on the unit cube in \ R^{n}[/itex]? What is the average distance of the vertices to the origin?&amp;amp;lt;br /&amp;amp;gt; &amp;amp;lt;br /&amp;amp;gt; First part I have \ 2^{n}[/itex], second part a square root of [ a summation from i = 1 to i = n of \ n^{2}[/itex]]&amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;gt; &amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;gt; But the last part stumps me...average distance? and writing a formula for this is a bit confusing too.

 

[mrandersdk]

for the first you should look at the gram-schmidt proces:

http://en.wikipedia.org/wiki/Gram-Schmidt_process

for the second, you are right about the 2^n. In general the distance from a vertice to the origin is given by

\frac{\sqrt{1^2+1^2+\dots+1^2}}{2} = \frac{\sqrt{n}}{2}

so the average must be

lim_{n\rightarrow \infty}\frac{1}{n}\sum_i^{n} \frac{\sqrt{i}}{2} = lim_{n\rightarrow \infty} \frac{1}{n}[\frac{\sqrt{1}+\sqrt{2}+\dots+\sqrt{n}}{2}] = lim_{n\rightarrow \infty} \frac{\sqrt{1}}{2n}+\frac{\sqrt{2}}{2n}+\dots +\frac{\sqrt{n}}{2n} = \infty

edited from:

lim_{n\rightarrow \infty}\frac{1}{n}\sum_i^{n} \frac{\sqrt{i}}{2} = lim_{n\rightarrow \infty} \frac{1}{n}[\frac{\sqrt{1}+\sqrt{2}+\dots+\sqrt{n}}{2}] = lim_{n\rightarrow \infty} \frac{\sqrt{1}}{2n}+\frac{\sqrt{2}}{2n}+\dots +\frac{\sqrt{n}}{2n} = 0

if the average is over dimension.

 

[mrandersdk]

ups, don't think the limit converge

 

[mrandersdk]

then I'm not sure what average you should calculate

 

[HallsofIvy]

I'm having trouble visualizing \ R^{4}[/itex](a domain of reals in four dimensions).<br /> <br /> 1. Describe a procedure in given 3 vectors, finds a fourth vector perpendicular to those three. Explain why we can use it in analogous fashion to the normal vector to a plane in \ R^{3}[/itex].&lt;br /&gt; &lt;br /&gt; Here, I&amp;#039;m thinking taking the normal vector of each vector and then adding the three normals would be sufficient, but I am not sure how this is analogous to the normal vector to a plane in the xyz system.

&lt;br /&gt; There is no such thing as &amp;quot;the&amp;quot; normal vector to a vector- any vector in the 3 dimensional &amp;quot;plane&amp;quot; normal to that vector. And even in 3 dimensions, adding normal vectors to two given vectors will not in general give you a vector normal to both. In 3 dimensions you have to take the cross product of the two given vectors. Can you thknk of a procedure analagous to that? &lt;br /&gt; &lt;br /&gt; &lt;blockquote data-attributes=&quot;&quot; data-quote=&quot;&quot; data-source=&quot;&quot; class=&quot;bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch&quot;&gt; &lt;div class=&quot;bbCodeBlock-content&quot;&gt; &lt;div class=&quot;bbCodeBlock-expandContent js-expandContent &quot;&gt; 2. How many vertices does the unit cube have in \ R^{n}[/itex] have? What is the furthest distance from the origin that one can be on the unit cube in \ R^{n}[/itex]? What is the average distance of the vertices to the origin?&amp;amp;lt;br /&amp;amp;gt; &amp;amp;lt;br /&amp;amp;gt; First part I have \ 2^{n}[/itex], second part a square root of [ a summation from i = 1 to i = n of \ n^{2}[/itex]]&amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;gt; &amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;gt; But the last part stumps me...average distance? and writing a formula for this is a bit confusing too. &lt;/div&gt; &lt;/div&gt; &lt;/blockquote&gt;&amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;gt; Assuming by &amp;amp;amp;amp;amp;quot;unit cube&amp;amp;amp;amp;amp;quot; they mean an n-dimensional &amp;amp;amp;amp;amp;quot;cube&amp;amp;amp;amp;amp;quot; with center at the origin and each side of length 1, then you have, NOT a &amp;amp;amp;amp;amp;quot; summation from i = 1 to i = n of n^2&amp;amp;amp;amp;amp;quot; but only a summation from 1 to n of 1 (the square of the side length= 1)- and that is just n. The diagonal distance between opposite vertices is \sqrt{n} and if the origin is at the center of the &amp;amp;amp;amp;amp;quot;cube&amp;amp;amp;amp;amp;quot; the distance you want is \sqrt{n}/2 (I see that mrandersdk has already given that). Now I guess you could divide that by n and sum to get an average but that does not converge!