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Adding Vectors

by odolwa99
Tags: adding, vectors
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odolwa99
#1
Jan13-13, 11:01 AM
P: 85
1. The problem statement, all variables and given/known data

Just a quick question. In the attached image, I can say [itex]\vec{am}=\vec{c}-\frac{1}{2}\vec{a}[/itex]. Although subtraction is not commutative, can I also say (relative strictly to vectors) that [itex]\vec{am}=-\frac{1}{2}\vec{a}+\vec{c}[/itex], considering [itex]\vec{am}=\frac{1}{2}\vec{ao}+\vec{c}[/itex]?

Many thanks.
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Mark44
#2
Jan13-13, 12:10 PM
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Quote Quote by odolwa99 View Post
1. The problem statement, all variables and given/known data

Just a quick question. In the attached image, I can say [itex]\vec{am}=\vec{c}-\frac{1}{2}\vec{a}[/itex]. Although subtraction is not commutative, can I also say (relative strictly to vectors) that [itex]\vec{am}=-\frac{1}{2}\vec{a}+\vec{c}[/itex], considering [itex]\vec{am}=\frac{1}{2}\vec{ao}+\vec{c}[/itex]?

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
#3
Jan13-13, 12:13 PM
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Quote Quote by odolwa99 View Post
1. The problem statement, all variables and given/known data

Just a quick question. In the attached image, I can say [itex]\vec{am}=\vec{c}-\frac{1}{2}\vec{a}[/itex]. Although subtraction is not commutative, can I also say (relative strictly to vectors) that [itex]\vec{am}=-\frac{1}{2}\vec{a}+\vec{c}[/itex], considering [itex]\vec{am}=\frac{1}{2}\vec{ao}+\vec{c}[/itex]?

Many thanks.
Saying that [itex]\ \vec{am}=\vec{c}-\frac{1}{2}\vec{a}\ [/itex] is essentially the same as saying [itex]\ \vec{am}=-\frac{1}{2}\vec{a}+\vec{c}\,,\ [/itex] because [itex]\ \vec{c}-\frac{1}{2}\vec{a}=\vec{c}+ \left(-\frac{1}{2}\vec{a}\right)\ [/itex] and vector addition is commutative.

odolwa99
#4
Jan13-13, 12:26 PM
P: 85
Adding Vectors

Great. Thank you very much.


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