MHB *IBV5 The vectors u, v are given by u = 3i + 5j, v = i – 2j

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The discussion revolves around finding scalars a and b such that a(u + v) = 8i + (b - 2)j, given the vectors u = 3i + 5j and v = i - 2j. The sum of the vectors is calculated as u + v = 4i + 3j. By setting up the equations 4a = 8 and 3a = b - 2, it is determined that a = 2 and b = 8. Participants clarify the definition of scalars as numbers, applicable in both real and complex contexts. The method discussed is noted as efficient for solving similar problems.
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The vectors $u, v$ are given by $u = 3i + 5j, v = i – 2j$
Find scalars $a, b$ such that $a(u + v) = 8i + (b – 2)j$

$(u+v)=4i+3j$
in order to get the $8i$ let $a=2$
then $2(4i+3j)=8i+6j$
in order to get $6j$ let $b=8$ then $(8-2)j=6j$
so $a=2$ and $b=8$

I am not sure of the precise definition of what scalar means here...at least with vectors
 
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vectors u,v are given by u=3i+5j,v=i–2j
Find scalars a,b such that a(u+v)=8i+(b–2)j

u + v = <4,3>

Then <4a,3a> = <8,(b-2)>

4a = 8

3a = b-2
 
karush said:
I am not sure of the precise definition of what scalar means here...at least with vectors
The definition is the same with everything; basically a number. Whether it's restricted to real or complex numbers is usually clear from context.
 
tkhunny said:
vectors u,v are given by u=3i+5j,v=i–2j
Find scalars a,b such that a(u+v)=8i+(b–2)j

u + v = <4,3>

Then <4a,3a> = <8,(b-2)>

4a = 8

3a = b-2

yes, that looks like a much better way to solve it. especially if it gets a lot more complicated.
 
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