The displacement vectors A and B both have magnitudes of 3.30 m. The direction of vector A is 16.6 degrees.
Find the magnitude of |A+B|, |A-B|, |B-A|, and |A-2B| graphically.
PLEASE HELP ;/

- anonymous

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- anonymous

each vector has a y and x component

- anonymous

i and j :) ?

- anonymous

well you're given a magnitude

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- anonymous

|dw:1359604762762:dw|

- anonymous

you there?

- anonymous

|dw:1359605259789:dw|

- anonymous

yeah, im here.. trying to understand

- anonymous

|dw:1359605406061:dw|
these need direction due to vectors .. it says A direction is 16.6 degrees sooooo

- anonymous

|dw:1359605476238:dw|

- anonymous

so in order to make sense of each vector you need to break them down into right triangles that show components of x and y i'll start with A|dw:1359605592468:dw|

- anonymous

now using trig functions you can find the components correct ? can you do that?

- anonymous

using sohcahtoa right?

- anonymous

A_x=sin(16.6)*A?

- anonymous

yes

- anonymous

i got 3.16 as side A_y..

- anonymous

so now we have
\[A_x=Acos(16.6)\]
\[A_y=Asin(16.6)\]

- anonymous

you had them switched around =]

- anonymous

and 0.95 as a_y.. A=3.30 right?

- anonymous

yep sounds correct

- anonymous

atleast according to my diagram lol

- anonymous

alright so the B is easy since there is only one component to it... which is the \[B_y\]

- anonymous

B=3.30 but theres no theta?

- anonymous

yes but it's a simple line on the y axis so it only has one component an x is what i meant =/

- anonymous

so you have
\[B_x=3.30\]

- anonymous

|dw:1359606759899:dw|

- anonymous

now you just add the like components to get 2 sides and the magnitude will be the square root of those squared total components

- anonymous

"like components" ? 8.1?

- anonymous

so B is 3.30 and for A do I add a_x and a_y and A?

- anonymous

well lets do it like my book says
\[R_x=A_x+B_x\]
\[R_y=A_y+B_y\]

- anonymous

\[|A+B|=\sqrt{R_x^2+R_y^2}\]

- anonymous

R_x=4.25
R_y= 3.16

- anonymous

|dw:1359607102751:dw|

- anonymous

5.3 m?

- anonymous

\[B_x=3.30\]
\[A_x=3.30cos(16.6)\]
\[A_y=3.30sin(16.6)\]

- anonymous

6.6 m?

- anonymous

\[R_x=6.46\]
\[R_y=.943\]
\[|R|=\sqrt{6.46^2+.943^2}\]

- anonymous

i'm getting 6.5

- anonymous

6.53 to be more exact

- anonymous

yeah i got that too now. i rounded wrong earlier.

- anonymous

so for A-B would i just subtract?

- anonymous

yes to do it easily i'm trying to remember what it looks like lol|dw:1359607683207:dw|

- anonymous

a-b is the diagonal from left to right if i remember right so you have to take A and move it infront of b like this

- anonymous

|dw:1359607791579:dw|

- anonymous

oh my lordy... i give up... ;/ my grade is good enough.. 96.3% on this assignment. I'm done with this question.. i still have to do a lab report and study for calc 2... thanks for the help though ;/

- anonymous

I'll go back later when i have more imte tomorrow and try to figure it out lol.

- anonymous

i looked at my book you have to face the vectors in certain ways so a-b says to face the vecotrs inward to eachother
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