Consider a triangle ABC formed by the vectorsa=(2,10,25) andb=(5,4,-2) attached at the point A. I've calculated the area of the triangle as $\displaystyle 40.5\sqrt{5} $ and now need to find the vectorh=AM, the height of the triangle [so AM is perpendicular to BC and M lies on the line BC]. I've tried using so many different formulae to try and findhor the third side of the triangle and nothing I get seems sensible. Can someone please give me some hints or suggestions on where to start so I can work through this problem? Thanks.