If $a + b + c = 6$ and $a^2 + b^2 + c^2 = 14$, find $ab + bc + ca$.

Using $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$ gives $36=14+2(ab+bc+ca)$. Therefore $ab+bc+ca=(36-14)/2=11$, so option A is correct. The cross-product sum is multiplied by $2$, so the difference $22$ must be divided by $2$.