Vector Subtraction: Is it Commutative?

[odolwa99]

Homework Statement

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.

 

[Mark44]

Homework Statement

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.

 

[SammyS]

Homework Statement

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.

 

[odolwa99]

Great. Thank you very much.