Since I’m at the topic of Vectors for most of my JC1 classes, I thought I share an interesting question regarding it. Yes, some students do tell me the questions aren’t really A’levels, but I think it does make you think and this is very important in learning.
Question: Can you place 5 points in three dimensional space such that they are pair wise equidistant?
The answer is no. Consider starting with three pairwise equidistant points with a common distant d between pairs. These are vertices of an equilateral triangle, and they will define a plane. If we add a 
point that is d units from each of the other three points, then it must be placed at a distance of latex \frac{\sqrt{3}d}{2}![away from the plane on the line normal to the plat that passes through the centre of the triangle. This will give us a tetrahedron. [caption id="attachment_2682" align="alignnone" width="256"]<a href="http://theculture.sg/wp-content/uploads/2015/12/HyperbolicTetrahedron_900.gif"><img class="size-medium wp-image-2682" src="http://theculture.sg/wp-content/uploads/2015/12/HyperbolicTetrahedron_900-256x300.gif" alt="Hyberbolice Tetrahedon Credits: wolfram.com" width="256" height="300" /></a> Hyberbolice Tetrahedon Credits: wolfram.com[/caption] You can consider adding a](http://theculture.sg/wp-content/ql-cache/quicklatex.com-4e64bbb2480fa34a29fb34f36c07aab9_l3.png "Rendered by QuickLaTeX.com")
latex 5^{th}$ point at another location that is d from each of the other four. But we will get the same result.
If you figured this out, or understand the explanation, here is a food for thought, in 4-dimensional space, the above will possible 🙂
