Question

Help with Vectors

Answer 1

For the following vector, v, find magnitude||u||, the unit vector with the same direction angle as v.

v= -1/2i-5/4j

Answer 2

\[\sqrt(-1/2)^2+(-5/4)^2=\frac{ \sqrt29 }{ 4}\]

Answer 3

u=\[\frac{ 2i \sqrt29 }{ 29 }-\frac{ 5j \sqrt29 }{ 29}\]

Answer 4

@amistre64 does this look right?

Answer 5

i followed your work till the magnitude part, where you calculated the magnitude of the vector which equals \[\sqrt{29}/4\] what did you do after that ?

Answer 6

if you want to find the magnitude of the unit vector then you divide the vector with your magnitude to get unit vector

Answer 7

not the magnitude of the unit vector lol, just the unit vector

Answer 8

Oh you divide the vecteor by the magnitude

Answer 9

so

Answer 10

\[\frac{ -1/2i-5/4j }{ \sqrt29/4 }\]

Answer 11

that is what i believe. However i would certainly ask @amistre64 to confirm this

Answer 12

-1/2i -5/4j

they use i j i use x y ... names a name

square add sqrt


1/4 + 25/16

29/16

sqrt(29)/4

Answer 13

divide it, yep its good

Answer 14

Thank you for checking it :D

Answer 15

-1/2i -5/4j


-2/4i -5/4j
----------
sqrt(29)/4


-2i -5j
-------
sqrt(29)

Answer 16

then multiply both sides be sqrt29 to get sqrt 29 on top and just 29 on bot

Answer 17

sqrt29/sqrt29*

Answer 18

coulda elongated the vector if we wanted to to get rid of the fractions

-2/4i -5/4j

-2i -5j is in the same direction just not unit

4 + 25 = 29; and we stil divide off sqrt(29)

Answer 19

well

1/sqrt(29) = sqrt(29)/29

Answer 20

interesting i did not know that

Answer 21

when ever in doubt, call @amistre64 to save the day :p