How Do You Calculate Vector Sums and Differences in Physics?

[tjbateh]

Homework Statement

Three vectors a , b , and c each have a magnitude of 42 m and lie in an xy plane. Their directions relative to the positive direction of the x-axis are 29°, 197°, and 314°, respectively. What are (a) the magnitude and (b) the angle of the vector a+ b + c, and (c) the magnitude and (d) the angle of a - b + c? What are (e) the magnitude and (f) the angle of a fourth vector d such that (a + b) - ( c+d ) = 0?

Homework Equations

The Attempt at a Solution

I do not how to approach this problem. Could someone please give me a few steps to try. Would be greatly appreciated!

 

[rl.bhat]

Draw the vectors according their positions. Resolve them along x-axis and y-axis.
Find the resultant of x and y components. And get the final answer using Pythagoras theorem.
Whenever the vector is negatived, reverse the vector and proceed.

 

[tjbateh]

I got A=36.73i + 20.36j
B= -19.07i + (-37.42j)
C= 29.18i + (-32.21j)

a+b+c= 46.84i - 29.27j

It was not correct? Do you see any mistakes?

 

[Hurkyl]

Do you see any mistakes?

We can't see the mistakes if you don't show your work...

 

[tjbateh]

Ok.

I did 42*cos(29)=36.73
42*sin(29)=20.36

42*cos(63) Got the 63 from 260-197...and that =19.07, which i made negative
42*sin(63)=37.42..also made negative

and..

42*cos(46)..Got the 46 from 360-314...and that = 29.18
42*sin(46)= I just did the calculation, now I'm get 30.21 instead of 32.31...other than that mistake, do you see anything else?

 

[rl.bhat]

[42*cos(63) Got the 63 from 260-197...and that =19.07, which i made negative]

This is wrong. It should be 270 - 197