Prove that the vectors a=3i+j-2k, b=-i+3j+4k, c=4i-2j-6k can form the sides of a triangle. Find the length of the medians of the triangle.
a-b=c. So, it's a triangle. Regarding median, I wonder how to approach. If the position vectors were given, we could do c-(a+b)/2. But how to proceed now?
Let $CC_1$ be a median of $\Delta ABC$.
Thus, $$CC_1=|\vec{CC_1}|=\left|\frac{1}{2}\left(\vec{CA}+\vec{CB}\right)\right|=$$ $$=\left|\frac{1}{2}(a-c+b-c)\right|=\left|\frac{1}{2}(a+b-2c)\right|=$$ $$=\left|\frac{1}{2}(3i+j-2k-i+3j+4k-2(4i-2j-6k))\right|=$$ $$=|-3i+4j+7k|=\sqrt{(-3)^2+4^2+7^2}=\sqrt{74}.$$
Lengths of other medians we can get by the similar way.