Vertices of an equilateral triangle Question: show that the following three points in 3D space A = <-2,4,0>, B = <1,2,-1> C = <-1,1,2> form the vertices of an equilateral triangle.
How do i approach this problem?
 A: Hint: A triangle is equilateral if and only if all its sidelengths are equal. We compute
$$\|A - B\| = \|\langle 3, 2, 1\rangle\| = \sqrt{3^2 + 2^2 + 1^2} = \sqrt{14}$$
$$\|A - C\| = \|\langle -1, 3, -2\rangle\| = ? $$
$$\|B - C\| = \|\langle 2, 1, -3\rangle\| = ? $$
A: Find the distance between all the pairs of points 
$$|AB|,|BC|,|CA|$$
and check if
$$|AB|=|BC|=|CA|$$
For example:
$$|A B| = \sqrt{(-2-1)^2 + (4-2)^2 + (0-(-1))^2} = \sqrt{14}$$
A: Apart for showing their length are the same, you will have to show they are not collinear. Construct a line equation with any two given points and check that the remaining point does not lie on the line. The other way is to check the cross product of AB and BC to make sure that it is not equal to 0. :)
A: Another, fancy approach: 
Calculate the directed vectors
$$\underline u:=\vec{AB}=B-A=(3,-2,-1)\;,\;\;\underline v:=\vec{AC}=C-A=(1,-3,2)$$
Now calculate the angle $\;\theta:=\angle BAC\;$ using the usual inner product
$$\cos\theta:=\frac{\underline u\cdot\underline v}{||\underline u||\;||\underline v||}=\frac{3+6-2}{\sqrt{14}\cdot\sqrt{14}}=\frac12\implies\theta=\arccos\frac12=\frac\pi3 (=60^\circ)$$
Well, do something similar for any other vertex you want (say, $\;\vec{BA}\,,\,\vec{BC}\;$) and check you get again the same angle, and a triangle with two angles of $\;60^\circ\;$ must be an equilateral one...
A: This would be the Grassmann algebra approach. This is a fairly simple problem and the main advantage of Grassmann algebra is that we can set up the objects and calculate with them in a natural manner.
<< GrassmannCalculus`
SetPreferences["Grassmann3Space", "Vector"]

The following defines the three vertices as points in 3-space:
vertexA = Origin + FreeBasis.{-2, 4, 0}
vertexB = Origin + FreeBasis.{1, 2, -1}
vertexC = Origin + FreeBasis.{-1, 1, 2} 

Giving:



The Measure routine gives the distance between any pair of points. We check if they are all equal.
Measure[vertexB - vertexA] == Measure[vertexC - vertexB] == 
 Measure[vertexA - vertexC] 
True

If we want the values:
{Measure[vertexB - vertexA], Measure[vertexC - vertexB], 
 Measure[vertexA - vertexC]} 

Giving:

