please help.. idk how to even figure this out so steps would be nice too. Find the specified vector or scalar. u = -4i + 1j and v = 4i + 1j; Find . ||u+v|| A. Sqrt34 B. 8 C. 5 D. 2

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please help.. idk how to even figure this out so steps would be nice too. Find the specified vector or scalar. u = -4i + 1j and v = 4i + 1j; Find . ||u+v|| A. Sqrt34 B. 8 C. 5 D. 2

Mathematics
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  • phi
first add the "corresponding" terms then find the length of the resulting vector (length is sqrt(sum of square of each term))
  • phi
for example the length of 3i+4j is \[ \sqrt{3^2+4^2} = \sqrt{9+16}= \sqrt{25} = 5\]
hey thank you for that a answer can you show me how you got got the length of the vector

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  • phi
what did you get for u+v?
wait i misunderstood i thought that was the same thing. im so confused
first add corresponding terms gives us... i+2j ?
idk how to find the length of a vector
  • phi
-4i + 1j 4i + 1j
  • phi
-4i + 4i is not i
its 0?
i dont understand the whole vector and scalar thing
  • phi
yes, the i's "go away" you are left with 2j
which is positive points?
  • phi
in 2-dimensions, you can think of the vector as pair of numbers for example, u= -4i + 1j (which can also be written <-4,1> in a graph, it looks like this: |dw:1443634973472:dw|
  • phi
and the length of the vector is the length of the line from (0,0) (the origin) to the point (-4,1) we use pythagoras to find its length
okay so U is a vector and length would be to the point.
oh okay i didnt see your message
  • phi
u+v = 2j (or 0i+2j, or (0,2) ) in a graph it looks like this |dw:1443635111360:dw|
|dw:1443635167339:dw|
  • phi
we could use pythagoras \[ \sqrt{0^2+2^2}= \sqrt{0+4}= \sqrt{4}= 2 \] but we can see the length is just 2
oh cool good so our drawings of vthis match okay so 0^2 because we have no i left and 2^2 becayse 2j
i understand thank you so much... can i ask what do i and j standfor why not use x and y? i got behind and i am trying to catch up on things as its obvious i am missing a gfew things in my knowledge
  • phi
The interesting thing is this same idea works for 3-D or even higher dimensions (though visualizing a vector with more than 3 components is beyond me)
  • phi
I'm not sure why people use i,j,k but it does not matter the idea is they are different "dimensions" for example, sideways and up/down. The other way people write vectors is as a "tuple" such as (1,2) or <1,2> where it is understood each number is the distance along each dimension
okay so vectors would be written as plot points or coordinates just signified as different dimensions thanks for the help on the problem and that little info there it caught me up a lil bit. medal and fan for you :)

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