need help with a calc III hw problem

[MrDudeMan]

i just need to know that i did this correctly...

i drew it in paint to avoid complications explaining it. its not hard, i just dont remember how to do it exactly. the best i could come up with is the resultant vector is 1.05i - 1.16j

any help would be appreciated

untitled.JPG

the question is simply find the sum of the vectors and the angle of the resultant from the x axis. this is in 2D. only x and y axis.


my answer:

||R|| = 1.56
theta = -47.85 degrees from +x

 

[dighn]

what's the question?!

and what are the axis on that diagram?

 

[ArmenK]

5i-8j+7.9cos30j-7.9sin30i=1.05i-1.1584j
||R||=(1.05^2+(-1.1584)^2)^.5=1.56345
theta=-arctan(1.1584/1.05)=-47.8101 degrees

 

[MrDudeMan]

Originally posted by: ArmenK
5i-7.9j+7.9cos30j-7.9sin30i

see if you get that same answer with my updated post

 

[MrDudeMan]

Originally posted by: ArmenK
5i-8j+7.9cos30j-7.9sin30i=1.05i-1.1584j
||R||=(1.05^2+(-1.1584)^2)^.5=1.56345
theta=-arctan(1.1584/1.05)=-47.8101 degrees

awesome

thanks

 

[Random Variable]

Just resolve the the vectors into their components and then add them up.

Rx= 5.00-7.90cos(60)= 1.05
Ry= 7.90sin(60)-8= -1.1583

R=[(1.05^2)+(-1.1584)^2)]^(1/2)= 1.5635
theta=arctan(1.1584/1.05)=47.810 degreees below the postive x-axis

 

[MrDudeMan]

Originally posted by: Vespasian
Just resolve the the vectors into their components and then add them up.

Rx= 5.00-7.90cos(60)= 1.05
Ry= 7.90sin(60)-8= -1.1583

R=[(1.05^2)+(-1.1584)^2)]^(1/2)= 1.5635
theta=arctan(1.1584/1.05)=47.810 degreees below the postive x-axis

cool. thats what i did, but i wasnt sure if it was right. just learning ya know =)

thx for the help vesp