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Mathematics 19 Online
OpenStudy (erinweeks):

Find the specified vector or scalar. u = -4i + 1j and v = 4i + 1j; Find ‖u+v‖.

OpenStudy (erinweeks):

A. √34 B. 8 C. 2 D. 5

OpenStudy (psymon):

Well do we know how to do the u+v part or should we go over that part, too?

OpenStudy (erinweeks):

go over it ..i dont know how to.

OpenStudy (psymon):

Alright. So you can see the vectors have an i part and a j part. When we add or subtract vectors, we combine the i parts and the j parts. So combining the i parts, we are doing -4i + 4i and for the j we have 1j + 1j. So this is just combining like terms to leave us with simply 2j. That make sense?

OpenStudy (erinweeks):

yes makes sense.

OpenStudy (psymon):

Alright, cool. Now the || || part means magnitude. It essentially just wants the length of the vector. Now we find the length the same way we would find the hypotenuse of a triangle or the distance between two points, with pythagorwan theorem/distance formula. If we recall, this is: \[c = \sqrt{a ^{2}+b ^{2}} \] Except this time we have i and j instead. Doesn't matter, process is the same. So now we do this: \[||u+v|| = \sqrt{(0)^{2}+(2)^{2}}\] I leave the i and the j out because we are only conncerned about the numbers. So if we do that calculation, what do we get?

OpenStudy (erinweeks):

you get \[\sqrt{4}\] ??

OpenStudy (psymon):

Yep, which reduces to 2. So your magnitude is just 2. Which makes sense. Before we took the magnitude, we just had 2j. Well, all that means is we go 0 units right and 2 units up |dw:1376622236102:dw| which this naturally has a length(magnitude) of 2 :3

OpenStudy (erinweeks):

okay .. than anything else.

OpenStudy (psymon):

No, that's all that was needed. We added u and v to get 2j, then did distance formula to see that our magnitude was just 2.

OpenStudy (erinweeks):

okay thought so makes sense(:

OpenStudy (psymon):

Okay, cool ^_^

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