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.
karush
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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.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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