Let ABCDEF be a regular hexagon. If
Let ABCDEF be a regular hexagon. If!AB = a and !BC = b, express each of the following as linear combinations of a and b: (a)!CD; (b)!DA; (c)!EB.
ab, bc are vectors
I can't read your notation there. But I'm sure you already know that all the vectors have to be the same length, and the angles are determined by the vertex angles of a hexagon.
yep
hmm, actually depending on how you label your vertices, AD and BE are diagonals, not sides and will have different lengths.
Probably best to start by drawing a regular hexagon and overlaying a coordinate grid with origin at a convenient vertex.
I'd probably do something like this: |dw:1347059834705:dw| I think that would be easier to get the vector components of everything you need.
A little basic geometry will get all the lengths and coordinates of vertices you need.
in the terms of a and b?
Sure. First start from that picture and get a and b in xi +yj vector component form, then once you know the distances and angles between any two vertices, you can convert that information into vectors.
ok, thanks
You're welcome. This is a fun one. I'm going to try to work it out. If you get an answer or something good along the way, let me know so I can check my own work.
ok, sounds good
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