## yajjnobee Group Title vector A has components Ax=1.30cm, Ay= 2.25cm; vector B has components Bx=4.10cm, By=-3.75cm. Find a. the components of the vector sum A+B; b. the magnitude and direction of vector A+B; c. the components of the vector difference B-A; d. the magnitude and direction of Vector B - A. 2 years ago 2 years ago

1. imranmeah91

A=<1.3,2.25> B=<4.1,-3.75> just add component by component A+B=<1.3+4.1,2.25-3.75> <5.4,-1.5> For magnitude $\sqrt{(5.4)^2+(1.5)^2}$ for direction $tan^{-1}(-1.5/5.4)$

2. yajjnobee

whar? we have different answers

3. imranmeah91

let me see what you do

4. yajjnobee

the formula to get the vector is $vector A= \sqrt{Ax ^{2}+ Ay^2}$ and $Vector B= \sqrt{Vx^{2} + Vy^{2}\} 5. imranmeah91 that will find you magnitude of vector A only 6. yajjnobee and vector b=\[\sqrt{Bx^2+ By^2}$

7. imranmeah91

oh ,you will add it afterward

8. yajjnobee

just add it? so I came up with vector A=2.6 and vector B=5.56 so vector A + vector B = 2.6 + 5.56=8.16?

9. imranmeah91

I think you are better off adding two vector component by component and then taking mag

10. yajjnobee

I think that part is on how to get the magnitude. right?

11. imranmeah91

magnitude of A+B

12. yajjnobee

yeah so the magnitude is $\sqrt{(5.4)^2 + (-1.5)^2}$?

13. imranmeah91

Yes

14. yajjnobee

will I do the same thing on part c and d?

15. yajjnobee

hey, ahmm, accordingly, when u get the magnitude, the formula for x-comp. is A cos $\Theta$ right so we need to find the angle before we get the magnitude?

16. imranmeah91

oh , if you are given magnitude and direction , yes

17. yajjnobee

i dunno what to do. what will be my answer on letter b. drivin me crazy.

18. yajjnobee

how can I get the value of x-component if the angle is not given. and it's impossible to compute the direction first because it needs the value for both x and y components