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Mathematics 11 Online
OpenStudy (nthenic_oftime):

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

OpenStudy (phi):

first add the "corresponding" terms then find the length of the resulting vector (length is sqrt(sum of square of each term))

OpenStudy (phi):

for example the length of 3i+4j is \[ \sqrt{3^2+4^2} = \sqrt{9+16}= \sqrt{25} = 5\]

OpenStudy (nthenic_oftime):

hey thank you for that a answer can you show me how you got got the length of the vector

OpenStudy (phi):

what did you get for u+v?

OpenStudy (nthenic_oftime):

wait i misunderstood i thought that was the same thing. im so confused

OpenStudy (nthenic_oftime):

first add corresponding terms gives us... i+2j ?

OpenStudy (nthenic_oftime):

idk how to find the length of a vector

OpenStudy (phi):

-4i + 1j 4i + 1j

OpenStudy (phi):

-4i + 4i is not i

OpenStudy (nthenic_oftime):

its 0?

OpenStudy (nthenic_oftime):

i dont understand the whole vector and scalar thing

OpenStudy (phi):

yes, the i's "go away" you are left with 2j

OpenStudy (nthenic_oftime):

which is positive points?

OpenStudy (phi):

in 2-dimensions, you can think of the vector as <x,y> 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|

OpenStudy (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

OpenStudy (nthenic_oftime):

okay so U is a vector and length would be to the point.

OpenStudy (nthenic_oftime):

oh okay i didnt see your message

OpenStudy (phi):

u+v = 2j (or 0i+2j, or (0,2) ) in a graph it looks like this |dw:1443635111360:dw|

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