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Mathematics 9 Online
OpenStudy (nthenic_oftime):

medal given for help please :) Find the specified vector or scalar. Show your work. u = -4i + 1j and v = 4i + 1j; Find ‖u+v‖.

OpenStudy (phi):

do you know how to find u+ v (by adding "corresponding components" ? what do you get ?

OpenStudy (nthenic_oftime):

@phi sorry i had to go for a bit... i assume i add like terms like you usually would in opperations but im confussed about why it has two lines is U+V the same as‖u+v‖

OpenStudy (nthenic_oftime):

is that what make it a scalar or vector? i think thats what im not understanding

OpenStudy (phi):

the first step is find u+v what did you get ?

OpenStudy (phi):

the 2nd part is find the "norm" of the vector you got from the first step. the norm or length is found using the dot product \[ | u |^2 = u \cdot u\] and then taking the square root

OpenStudy (nthenic_oftime):

okay so for the U+V i got 2J is that correct? @phi

OpenStudy (phi):

yes the norm is the length of the vector 2j which by inspection is 2 but you can use the formula. write the vector as <0,2> and find the length squared: <0,2> dot <0,2> = 0*0+2*2 = 4 now take the square root to find the length: sqr(4) = 2

OpenStudy (nthenic_oftime):

so essentially i am looking for the length of the vector squared multipliesd the vector to square it and the square root so for this im right in saying that ‖u+v‖ = SQRT(4) = 2?

OpenStudy (phi):

you use a dot product, which means you multiply corresponding components then add for example for vectors <a,b> and <c,d> you would do a*c+b*d if you do that for the same vector: <a,b> dot <a,b> you get a^2 + b^2 and that is the "length squared"

OpenStudy (phi):

or, if you plot a vector on the i-axis , j-axis |dw:1459202536373:dw|

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