# triangles represented using vectors

• Apr 21st 2010, 08:05 AM
calliope
triangles represented using vectors
Consider a triangle ABC formed by the vectors a=(2,10,25) and b=(5,4,-2) attached at the point A. I've calculated the area of the triangle as $40.5\sqrt{5}$ and now need to find the vector h=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 find h or 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.
• Apr 23rd 2010, 05:06 PM
hollywood
The line going through $A$ and $B$ is just $A+\gamma(B-A)$, where $\gamma$ is a real number. So $\gamma=0$ gives point $A$ and $\gamma=1$ gives point $B$.

The altitude $AM$ of the triangle has to be perpendicular to that line, so

$M\cdot(B-A)=(A+\gamma(B-A))\cdot(B-A)=0$

which you should be able to solve for $\gamma$, and then find $M$.

Post again in this thread if you're still having trouble.

- Hollywood