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Mathematics 21 Online
OpenStudy (anonymous):

Find the magnitude and direction of vector XY for X(1, -4) and Y(6, 3). Show all work.

OpenStudy (anonymous):

magnitude is root(74)

OpenStudy (anonymous):

direction arctan (1.4)

OpenStudy (anonymous):

how do you get that? I dont understand.

OpenStudy (anonymous):

the magnitude is the distance between the 2 pts

OpenStudy (angela210793):

\[=\sqrt{(x2-x1)^2+(y2-y1)^2}\]

OpenStudy (anonymous):

whts the vector XY?

OpenStudy (anonymous):

@him1618 I dont know. Thats all I was given. @angela210793 isnt that the distance formula? if so I understand!

OpenStudy (angela210793):

XY=(6-1)=(5) (4+4) (8)

OpenStudy (anonymous):

vector XY can be obtained by X - Y..understand?

OpenStudy (anonymous):

@angela its 5,7

OpenStudy (angela210793):

yea tht's the distance and @Him aren't wht i wrote the coordinates of XY?

OpenStudy (angela210793):

wht's 5.7?

OpenStudy (anonymous):

XY is a vector angela it doesnt have coordinates im saying the vector XY is 5,7..u wrote 5,8

OpenStudy (angela210793):

yea ur right is 5;7(typing mistake) and as far as i know a vector does have coordinates...how come it hasn't?

OpenStudy (anonymous):

ur confusing it...u mean to say 5 and 7 are the coordinates of a vector..they are called COMPONENTS of a vector

OpenStudy (anonymous):

ONLY in the case of a POSITION VECTOR, the components of a vector become the coordinates of a point..understand?

OpenStudy (anonymous):

\[\tan x=7/5\] \[x = \tan^{-1} (7/5)\] \[\tan x=8/5\] \[x = \tan^{-1} (8/5)\] is this the right format?

OpenStudy (angela210793):

Hmmm...:(:(:( we call them coordinates as well :(:( Sorry my fault :( :)

OpenStudy (anonymous):

im looking at finding the direction, btw

OpenStudy (anonymous):

the first ones right @angela: no prob, my fault

OpenStudy (anonymous):

Thank You.

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