by odolwa99
 P: 85 1. The problem statement, all variables and given/known data Just a quick question. In the attached image, I can say $\vec{am}=\vec{c}-\frac{1}{2}\vec{a}$. Although subtraction is not commutative, can I also say (relative strictly to vectors) that $\vec{am}=-\frac{1}{2}\vec{a}+\vec{c}$, considering $\vec{am}=\frac{1}{2}\vec{ao}+\vec{c}$? Many thanks. Attached Thumbnails
Mentor
P: 21,314
 Quote by odolwa99 1. The problem statement, all variables and given/known data Just a quick question. In the attached image, I can say $\vec{am}=\vec{c}-\frac{1}{2}\vec{a}$. Although subtraction is not commutative, can I also say (relative strictly to vectors) that $\vec{am}=-\frac{1}{2}\vec{a}+\vec{c}$, considering $\vec{am}=\frac{1}{2}\vec{ao}+\vec{c}$? Many thanks.
Subtraction isn't commutative, as you noted, but addition is, and that's really what you're doing. a - b = a + (-b), which is the same as -b + a.
Emeritus
 Quote by odolwa99 1. The problem statement, all variables and given/known data Just a quick question. In the attached image, I can say $\vec{am}=\vec{c}-\frac{1}{2}\vec{a}$. Although subtraction is not commutative, can I also say (relative strictly to vectors) that $\vec{am}=-\frac{1}{2}\vec{a}+\vec{c}$, considering $\vec{am}=\frac{1}{2}\vec{ao}+\vec{c}$? Many thanks.
Saying that $\ \vec{am}=\vec{c}-\frac{1}{2}\vec{a}\$ is essentially the same as saying $\ \vec{am}=-\frac{1}{2}\vec{a}+\vec{c}\,,\$ because $\ \vec{c}-\frac{1}{2}\vec{a}=\vec{c}+ \left(-\frac{1}{2}\vec{a}\right)\$ and vector addition is commutative.