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?

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...
